{"id":20959,"date":"2022-08-17T10:13:09","date_gmt":"2022-08-17T04:43:09","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=20959"},"modified":"2022-08-17T10:35:50","modified_gmt":"2022-08-17T05:05:50","slug":"inter-2nd-year-maths-2b-system-of-circles-important-questions","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/inter-2nd-year-maths-2b-system-of-circles-important-questions\/","title":{"rendered":"Inter 2nd Year Maths 2B System of Circles Important Questions"},"content":{"rendered":"

Students get through Maths 2B Important Questions<\/a> Inter 2nd Year Maths 2B System of Circles Important Questions which are most likely to be asked in the exam.<\/p>\n

Intermediate 2nd Year Maths 2B System of Circles Important Questions<\/h2>\n

Question 1.
\nx2<\/sup> + y2<\/sup> + 4x + 8 = 0, x2<\/sup> + y2<\/sup> – 16y + k = 0 [A.P. & T.S. Mar. 16]
\nSolution:
\ng1<\/sub> = 2; f1<\/sub> = 0; c1<\/sub> = 8
\ng2<\/sub> = 0; f2<\/sub> = – 4; c2<\/sub> = k
\nTwo circles are said to be orthogonal
\n2g1<\/sub>g2<\/sub> + 2f1<\/sub>f2<\/sub> = c1<\/sub> + c2<\/sub>
\n2(2) (0) + 2(0) (-8) = 8 + k
\n0 + 0 = 8 + k
\n\u21d2 k = -8<\/p>\n

\"Inter<\/p>\n

Question 2.
\n(x – a)2<\/sup> + (y – b)2<\/sup> = c2<\/sup>, (x – b)2<\/sup> + (y – a)2<\/sup> = c2<\/sup> (a \u2260 b) [A.P. Mar. 15]
\nSolution:
\n(x2<\/sup> + y2<\/sup> – 2xa – 2yb – c2<\/sup>)
\n– (x2<\/sup> + y2<\/sup> – 2xb – 2ya – c2<\/sup>) = 0
\n– 2x (a – b) – 2y(b – a) = 0
\n(or) x – y = 0<\/p>\n

Question 3.
\nFind the angle between the circles given by the equations. [T.S. Mar.17]
\ni) x2<\/sup> + y2<\/sup> – 12x – 6y + 41 = 0, x2<\/sup> + y2<\/sup> + 4x + 6y – 59 = 0.
\nSolution:
\nC1<\/sub> = (6, 3)
\nr1<\/sub> =(36 + 9 – 41)1\/2<\/sup>
\nr1<\/sub> = 2<\/p>\n

C2<\/sub> = (-2, -3)
\nr2<\/sub> =(4 + 9 + 59)1\/2<\/sup>
\nr2<\/sub> = (72)1\/2<\/sup> = 6\\(\\sqrt{2}\\)<\/p>\n

C1<\/sub>C2<\/sub> = d = \\(\\sqrt{(6+2)^{2}+(3+3)^{2}}\\)
\n= \\(\\sqrt{64+36}\\) = 10
\ncos \u03b8 = \\(\\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\\)
\n= \\(\\frac{100-4-72}{2 \\times 2 \\cdot \\sqrt{72}}=\\frac{24}{4 \\times 6 \\sqrt{2}}=\\frac{1}{\\sqrt{2}}\\)
\n\u03b8 = 45\u00b0<\/p>\n

\"Inter<\/p>\n

Question 4.
\nFind the equation of the circle which cuts the circles x2<\/sup> + y2<\/sup> – 4x – 6y + 11 = 0 and x2<\/sup> + y2<\/sup> – 10x – 4y + 21 = 0 orthogonally and has the diameter along the straight line 2x + 3y = 7. [A.P. Mar. 16; May 07]
\nSolution:
\nLet circle be x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 ……………….. (i)
\nOrthogonal to circle
\n2g (-2) +2f(-3) = 11 + c ……………………. (ii)
\n2g (-5) + 2f(-2) = 21 + c ……………………. (iii)
\nSubtracting it we get
\n-6g + 2f = 10 ……………………… (iv)
\nCircles centre is on 2x + 3y = 7
\n\u2234 -2g – 3f = 7 ……………………. (v)
\nSolving (iv) and (y)
\nf = -1, g = -2, c = 3
\nEquation of circle be x2<\/sup> + y2<\/sup> – 4x – 2y + 3 = 0<\/p>\n

Question 5.
\nShow that the angle between the circles x2<\/sup> + y2<\/sup> = a2<\/sup>, x2<\/sup> + y2<\/sup> = ax + ay is \\(\\frac{3\\pi}{4}\\) [Mar. 14]
\nSolution:
\nEquations of the circles are
\nS \u2261 x2<\/sup> + y2<\/sup> – a2<\/sup> = 0
\nS’ \u2261 x2<\/sup> + y2<\/sup> – ax – ay = 0
\nC1<\/sub> (0, 0), C2<\/sub> (\\(\\frac{a}{2}\\), \\(\\frac{a}{2}\\))
\n\u2234 C1<\/sub>C2<\/sub>2<\/sup> = (0 – \\(\\frac{a}{2}\\))2<\/sup> + (0 – \\(\\frac{a}{2}\\))2<\/sup>
\n\"Inter<\/p>\n

Question 6.
\nIf x + y = 3 \u00a1s the equation of the chord AB of the circle x2<\/sup> + y2<\/sup> – 2x + 4y – 8 = 0, find the equation of the circle having \\(\\overline{\\mathrm{AB}}\\) as diameter. [A.P. Mar. 15]
\nSolution:
\nRequired equation of circle passing through
\nintersection S = 0 and L = 0 is S + \u03bbL = 0
\n(x2<\/sup> + y2<\/sup> – 2x + 4y – 8) + \u03bb(x + y – 3) = 0
\n(x2<\/sup> + y2<\/sup> + x(-2 + \u03bb)+ y(4 + \u03bb) – 8 – 3\u03bb = 0 ………………. (i)
\nx2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 ……………………. (ii)
\nComparing (i) and (ii) we get
\ng = \\(\\frac{(-2+\\lambda)}{2}\\), f = \\(\\frac{(4+\\lambda)}{2}\\)
\nCentre lies on x + y = 3
\n\u2234 \\(-\\left(\\frac{-2+\\lambda}{2}\\right)-\\left(\\frac{4+\\lambda}{2}\\right)\\) = 3
\n2 – \u03bb – 4 – \u03bb = 6
\n-2\u03bb = 8 \u21d2 \u03bb = – 4
\nRequired equation of circle be
\n(x2<\/sup> + y2<\/sup> – 2x + 4y – 8) – 4(x + y – 3) = 0
\nx2<\/sup> + y2<\/sup> – 6x + 4 = 0<\/p>\n

\"Inter<\/p>\n

Question 7.
\nIf two circles x2<\/sup> + y2<\/sup> + 2gx + 2fy = 0 and x2<\/sup> + y2<\/sup> + 2g’x + 2f’y = 0 touch each other then show that f’g = fg’. [T.S. Mar. 16]
\nSolution:
\nC1<\/sub> = (-g, -f)
\nr1<\/sub> = \\(\\sqrt{g^{2}+f^{2}}\\)
\nC2<\/sub> = (-g-1<\/sup>, -f-1<\/sup>)
\nr2<\/sub> = \\(\\sqrt{g^{\\prime 2}+r^{\\prime 2}}\\)
\nC1<\/sub>C2<\/sub> = r1<\/sub> + r2<\/sub>
\n(C1<\/sub>C2<\/sub>)2<\/sup> = (r1<\/sub>r2<\/sub>)2<\/sup>
\n(g’ – g)2<\/sup> + (f’ – f)2<\/sup> = g2<\/sup> + f2<\/sup> + g’2<\/sup> + f’2<\/sup> + \\(2 \\sqrt{g^{2}+f^{2}} \\sqrt{g^{2}+f^{1^{2}}}\\)
\n-2(gg’ + ff’) = 2{g2<\/sup>g’2<\/sup> + g2<\/sup>f’2<\/sup> + f2<\/sup>g’2<\/sup>}1\/2<\/sup>
\nSquaring again
\n(gg’ + ff’)2<\/sup> = g2<\/sup>g’2<\/sup> + f2<\/sup>f’2<\/sup> + g2<\/sup>f’2<\/sup> + g’2<\/sup>f2<\/sup>
\ng2<\/sup>g’2<\/sup> + f2<\/sup>f’2<\/sup> + 2gg’ff’ = g2<\/sup>g’2<\/sup> + f2<\/sup>f’2<\/sup> + g2<\/sup>f’2<\/sup> + g’2<\/sup>f’2<\/sup>
\n2gg’ff’ = g2<\/sup>f’2<\/sup> + f2<\/sup>g’2<\/sup>
\n\u21d2 g2<\/sup>f’2<\/sup> + g’2<\/sup>f’2<\/sup> – 2gg’ff’ = 0
\n(or) (gf’ – fg’)2<\/sup> = 0 (or) gf’ = fg’<\/p>\n

Question 8.
\nShow that the circles
\nS \u2261 x2<\/sup> + y2<\/sup> – 2x – 4y – 20 = 0 …………………… (1)
\nand S \u2261 x2<\/sup> + y2<\/sup> + 6x + 2y – 90 = 0 …………………(2)
\ntouch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
\nSolution:
\nLet C1<\/sub>, C2<\/sub> be the centres and r1<\/sub>, r2<\/sub> be the radii of the given circles (1) and (2). Then
\nC1<\/sub> = (1, 2); C2<\/sub> = (-3, -1); r1<\/sub> = 5; r2<\/sub> = 10.
\nC1<\/sub>C2<\/sub> = distance between the centres = 5
\n|r1<\/sub> – r2<\/sub>| = |5 – 10| = 5 = C1<\/sub>C2<\/sub>
\n\u2234 The given two circles touch internally. In
\nthis case, the common tangent is nothing but the radical axis. Therefore its equation is S – S’ = 0.
\ni.e., 4x + 3y – 35 = 0
\nNow we find the point of contact. The point of contact divides \\(\\overline{\\mathrm{C}_{1} \\mathrm{C}_{2}}\\) in the ratio 5 : 10
\ni.e., 1: 2 (externally)
\n\u2234 Point of contact
\n= \\(\\left(\\frac{(1)(-3)-2(1)}{1-2}, \\frac{(1)(-1)-2(2)}{1-2}\\right)\\)
\n= (5, 5)<\/p>\n

\"Inter<\/p>\n

Question 9.
\nFind the angle between the circles x2<\/sup> + y2<\/sup> + 4x – 14y + 28 = 0 and x2<\/sup> + y2<\/sup> + 4x – 5 = 0
\nSolution:
\nEquations of the given circles are .
\nx2<\/sup> + y2<\/sup> + 4x – 14y + 28 = 0
\nx2<\/sup> + y2<\/sup> + 4x – 5 = 0
\nCentres are C1<\/sub> (-2, 7), C2<\/sub> (-2, 0)
\nC1<\/sub>C2<\/sub> = \\(\\sqrt{(-2+2)^{2}+(7-0)^{2}}\\)
\n= \\(\\sqrt{0+49}\\) = 7
\nr1<\/sub> = \\(\\sqrt{4+49-28}\\) = \\(\\sqrt{25}\\) = 5
\nr2<\/sub> = \\(\\sqrt{4+5}\\) = \\(\\sqrt{9}\\) = 3
\nIf \u03b8 is the angle between the given circles,
\nthen cos \u03b8 = \\(\\frac{d^{2}-r_{1}^{2}-r_{2}^{2}}{2 r_{1} r_{2}}\\)
\ncos \u03b8 = \\(\\frac{49-25-9}{2(5)(3)}\\) = \\(\\frac{15}{2.5 .3}\\) = \\(\\frac{1}{2}\\) = cos 60\u00b0
\nAngle between the circles = \u03b8 = \\(\\frac{\\pi}{3}\\)<\/p>\n

Question 10.
\nIf the angle between the circles x2<\/sup> + y2<\/sup> – 12x – 6y + 41 = 0 and x2<\/sup> + y2<\/sup> + kx + 6y – 59 = 0 is 45\u00b0 find k.
\nSolution:
\nSuppose \u03b8 is the angle between the circles
\nx2<\/sup> + y2<\/sup> – 12x – 6y + 41 = 0
\nand x2<\/sup> + y2<\/sup> + kx + 6y – 59 = 0
\ng1<\/sub> = -6, f1<\/sub> = -3, c1<\/sub> = 41,
\ng2<\/sub> = \\(\\frac{k}{2}\\), f2<\/sub> = 3, c2<\/sub> = -59
\n\"Inter
\n\\(\\frac{1}{\\sqrt{2}}=\\frac{6 k}{4 \\cdot \\sqrt{\\frac{k^{2}}{4}+68}}\\)
\nSquaring and cross – multiplying
\n4(\\(\\frac{k^{2}}{4}\\) + 68) = 18k2<\/sup>
\n\\(\\frac{2\\left[k^{2}+272\\right]}{4}\\) = 9k2<\/sup>
\nk2<\/sup> + 272 = 18 k2<\/sup>
\n17k2<\/sup> = 272
\nk2<\/sup> = \\(\\frac{272}{17}\\)
\nk2<\/sup> = 16
\nk = \u00b14.<\/p>\n

Question 11.
\nFind the equation of the circle which passes through (1, 1) and cuts orthogonally each of the circles.
\nx2<\/sup> + y2<\/sup> – 8x – 2y + 16 = 0 and …………….. (1)
\nx2<\/sup> + y2<\/sup> – 4x – 4y – 1 = 0. ………………. (2)
\nSolution:
\nLet the equation of the required circle be
\nx2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 ……………. (3)
\nThen the circle (3) is orthogonal to (1) and (2).
\n\u2234 By applying the condition of orthogonality give in x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 we get
\n2g(-4) + 2f(-1) = c + 16 and …………………. (4)
\n2g (-2) + 2f(-2) = c – 1 ……………….. (5).
\nGiven that the circle (3) is passing through (1, 1)
\n\u2234 12<\/sup> + 12<\/sup> + 2g(1) + 2f(1) + c = 0
\n2g + 2f + c + 2 = 0 ………………….. (6)
\nSolving (4), (5) and (6) for g, f and c, we get
\ng = –\\(\\frac{7}{3}\\), f = \\(\\frac{23}{6}\\), c = -5
\nThus the equation of the required circle is
\n3(x2<\/sup> + y2<\/sup>) – 14x + 23y – 15 = 0.<\/p>\n

\"Inter<\/p>\n

Question 12.
\nFind the equation of the circle which is orthogonal to each of the following three circles.
\nx2<\/sup> + y2<\/sup> + 2x + 17y + 4 = 0 ………………… (1)
\nx2<\/sup> + y2<\/sup> + 7x + 6y + 11 = 0 ………………. (2)
\nand x2<\/sup> + y2<\/sup> – x + 22y + 3 = 0 ………………… (3)
\nSolution:
\nLet the equation of the required circle be
\nx2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0 ……………… (4)
\nSince this circle is orthogonal to (1), (2) and (3). by applying the condition of orthogonality given in x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0
\nwe have
\n2(g)(1) + 2(f) (\\(\\frac{17}{2}\\)) = c + 4 ……………….. (5)
\n2(g) (\\(\\frac{7}{2}\\)) + 2(f)(3) = c + 11 ………………. (6)
\nand 2(g) (-\\(\\frac{1}{2}\\)) + 2(f)(11) = c + 3 ………………….. (7)
\nSolving (5), (6) and (7) for g, f, c we get
\ng = -3, f = -2 and c = -44
\nThus the equation of the required circle is
\nx2<\/sup> + y2<\/sup> – 6x – 4y – 44 = 0.<\/p>\n

Question 13.
\nIf the straight line is represented by
\nx cos \u03b1 + y sin \u03b1 = p ………………….. (1)
\nintersects the circle
\nx2<\/sup> + y2<\/sup> = a2<\/sup> ……………… (2)
\nat the points A and B, then show that the equation of the circle with \\(\\overline{\\mathrm{AB}}\\) as diameter is(x2<\/sup> + y2<\/sup> – a2<\/sup>) – 2p(x cos \u03b1 + y sin \u03b1 – p) = 0.
\nSolution:
\nThe equation of the circle passing through the points A and B is S \u2261 x2<\/sup> + y2<\/sup> + 2gx + 2fy + c = 0
\n(x2<\/sup> + y2<\/sup> – a2<\/sup>) + \u03bb(x cos \u03b1 + y sin \u03b1 – p) = 0 ……………… (3)
\nThe centre of this circle is
\n\\(\\left(-\\frac{\\lambda \\cos \\alpha}{2},-\\frac{\\lambda \\sin \\alpha}{2}\\right)\\)
\nIf the circle given by (3) has \\(\\overline{\\mathrm{AB}}\\) as diameter then the centre of it must lie on (1)
\n\u2234 \\(-\\frac{\\lambda \\cos \\alpha}{2}\\) (cos \u03b1) – \\(\\frac{\\lambda \\sin \\alpha}{2}\\) (sin \u03b1) = p
\ni.e., –\\(\\frac{\\lambda}{2}\\) (cos2<\/sup> \u03b1 + sin2<\/sup> \u03b1) p
\ni.e., \u03bb = -2p
\nHence the equation of the required circle is
\n(x2<\/sup> + y2<\/sup> – a2<\/sup>) – 2p(x cos \u03b1 + y sin \u03b1 – p) = 0.<\/p>\n

\"Inter<\/p>\n

Question 14.
\nFind the equation of the circle passing through the points of intersection of the circles.
\nx2<\/sup> + y2<\/sup> – 8x – 6y + 21 = 0 ……………….. (1)
\nx2<\/sup> + y2<\/sup> – 2x – 15 = 0 ……………….. (2)
\nand (1, 2).
\nSolution:
\nThe equation of circle passing through the points of intersection of (1) arid (2) is
\n(x2<\/sup> + y2<\/sup> – 8x – 6y + 21) + \u03bb (x2<\/sup> + y2<\/sup> – 2x – 15) = 0 ……………….. (3)
\nIf it passes through (1, 2). we obtain
\n(1 + 4 – 8 – 12 + 21) + \u03bb(1 + 4 – 2 – 15) = 0
\ni.e., 6 + \u03bb(-12) = 0
\ni.e., \u03bb = \\(\\frac{1}{2}\\)
\nHence the equation of the required circle is
\n(x2<\/sup> + y2<\/sup> – 8x – 6y + 21) + \\(\\frac{1}{2}\\) (x2<\/sup> + y2<\/sup> – 2x – 15) = 0
\ni.e., 3(x2<\/sup> + y2<\/sup>) – 18x – 12y + 27 = 0.<\/p>\n

Question 15.
\nLet us find the equation the radical axis of the circles S \u2261 x2<\/sup> + y2<\/sup> – 5x + 6y + 12 = 0 and S’ \u2261 x2<\/sup> + y2<\/sup> + 6x – 4y – 14 = 0
\nSolution:
\nThe given equations of circles are in general form. Therefore their radical axis is (S – S’ = 0)
\ni.e., 11x – 10y – 26 = 0<\/p>\n

Question 16.
\nLet us find the equation of the radical axis of the circles
\n2x2<\/sup> + 2y2<\/sup> + 3x + 6y – 5 = 0 ………………….. (1)
\nand 3x2<\/sup> + 3y2<\/sup> – 7x + 8y – 11 = 0 ……………….. (2)
\nSolution:
\nHence the given equations are not in general form, we get :
\nx2<\/sup> + y2<\/sup> + \\(\\frac{3}{2}\\)x + 3y – \\(\\frac{5}{2}\\) = 0 and
\nx2<\/sup> + y2<\/sup> – \\(\\frac{7}{3}\\)x + \\(\\frac{8}{3}\\) y – \\(\\frac{11}{3}\\) = 0
\nNow the radical axis equation of given circles is
\n(\\(\\frac{3}{2}\\) + \\(\\frac{7}{3}\\))x + (3 – \\(\\frac{8}{3}\\))y + (-\\(\\frac{5}{2}\\) + \\(\\frac{11}{3}\\)) = 0
\ni.e., 23x + 2y + 7 = 0<\/p>\n

\"Inter<\/p>\n

Question 17.
\nLet us find the radical centre of the circles
\nx2<\/sup> + y2<\/sup> – 2x + 6y = 0 ………………… (1)
\nx2<\/sup> + y2<\/sup> – 4x – 2y + 6 = 0 …………………. (2)
\nand x2<\/sup> + y2<\/sup> – 12x + 2y + 3 = 0 ………………. (3)
\nSolution:
\nThe radical axis of (1) and (2) and (3)
\nx + 4y – 3 = 0 ………………… (4)
\n8x – 4y + 3 = 0 …………………. (5)
\n10x + 4y – 3 = 0 ……………… (6)
\nSolving (4) and (5) for the point of intersection we get (0, \\(\\frac{3}{4}\\)) which is the required radical centre. Observe that the co-ordinates of this point satisfies (6) also.<\/p>\n

Question 18.
\nFind the equation and length of the common chord of the two circles
\nS \u2261 x2<\/sup> + y2<\/sup> + 3x + 5y + 4 = 0
\nand S’ \u2261 x2<\/sup> + y2<\/sup> + 5x + 3y + 4 = 0
\nSolution:
\nEquations of the given circles are
\nS \u2261 x2<\/sup> + y2<\/sup> + 3x + 5y + 4 = 0 ………………… (1)
\nS’ \u2261 x2<\/sup> + y2<\/sup> + 5x + 3y + 4 ………………. (2)
\nEquations of the common chord is S – S’ = 0
\n-2x + 2y = 0
\nL \u2261 x – y = 0 …………….. (3)
\n\"Inter<\/p>\n

Question 19.
\nShow that the circles
\nS \u2261 x2<\/sup> + y2<\/sup> – 2x – 4y – 20 = 0 ……………….. (1)
\nand S’ \u2261 x2<\/sup> + y2<\/sup> + 6x + 2y – 90 = 0 ……………. (2)
\ntouch each other internally. Find their point of contact and the equation of common tangent. [T.S. Mar. 15]
\nSolution:
\nLet C1<\/sub>, C2<\/sub> be the centres and r1<\/sub>, r2<\/sub> be the radii of the given circles (1) and (2). Then
\nC1<\/sub> = (1, 2); C2<\/sub> = (-3, -1); r1<\/sub> = 5; r2<\/sub> = 10
\nC1<\/sub>C2<\/sub> = distance between the centres = 5
\n|r1<\/sub> – r2<\/sub>| = |5 – 10| = 5 = C1<\/sub>C2<\/sub>
\n\u2234 The given two circles touch internally. In this case, the.common tangent is nothing but the radical axis. Therefore its equation is
\nS – S’ = 0
\ni.e., 4x + 3y – 35 = 0
\nNow we find the point of contact. The point of contact divides in the ratio 5 : 10 i.e., 1 : 2 (externally)
\n\u2234 Point, of contact
\n= \\(\\left(\\frac{(1)(-3)-2(1)}{1-2}, \\frac{(1)(-1)-2(2)}{1-2}\\right)\\)
\n= (5, 5).<\/p>\n

\"Inter<\/p>\n

Question 20.
\nFind the equation of the circle whose diameter is the common chord of the circles
\nS \u2261 x2<\/sup> + y2<\/sup> + 2x + 3y + 1 = 0 ……………….. (1)
\nand S’ \u2261 x2<\/sup> + y2<\/sup> + 4x + 3y + 2 = 0 ……………….. (2)
\nSolution:
\nHere the common chord is the radical axis of (1) and (2). The equation of the radical axis is S – S’ = 0.
\ni.e., 2x + 1 = 0 …………………… (3)
\nThe equation of any circle passing through
\nthe points of intersection of (1) and (3) is (S + \u03bbL = 0)
\n(x2<\/sup> + y2<\/sup> + 2x + 3y + 1) + \u03bb(2x + 1) = 0
\nx2<\/sup> + y2<\/sup> + 2(\u03bb + 1)x + 3y + (1 + \u03bb) = 0 ……………………. (4)
\nThe centre of this circle is (-(\u03bb + 1), \\(\\frac{3}{2}\\)).
\nFor the circle (4), 2x + 1 = 0 is one chord. This chord will be a diameter of the circle (4) if the centre of (4) lies on (3).
\n\u2234 2{-(\u03bb + 1)} + 1 = 0
\n\u21d2 \u03bb = –\\(\\frac{1}{2}\\)
\nThus equation of the circle whose diameter is the common chord (1) and (2)
\n(put \u03bb = \\(\\frac{1}{2}\\) in equation (4))
\n2(x2<\/sup> + y2<\/sup>) + 2x + 6y + 1 = 0<\/p>\n

Question 21.
\nLet us find the equation of a circle which cuts each of the following circles orthogonally
\nS’ \u2261 x2<\/sup> + y2<\/sup> + 3x + 2y + 1 = 0 ………………… (1)
\nS” \u2261 x2<\/sup> + y2<\/sup> – x + 6y + 5 = 0 …………….. (2)
\nand S” \u2261 x2<\/sup> + y2<\/sup> + 5x – 8y + 15 = 0 ……………………. (3)
\nSolution:
\nThe centre of the required circle is radical centre of (1), (2) and (3) and the radius is the length of the tangent from this point to any one of the given three circles. First we shall find the radical centre. For, the radical axis of (1) and (2) is
\nx – y = 1 ………………… (4)
\nand the radical axis of (2) and (3) is
\n3x – 7y = -5 …………………… (5)
\nThe point of intersection (3, 2) of (4) and (5) is the radical centre of the circles (1), (2) and (3). The length of tangent from (3. 2) to the circle (1)
\n= \\(\\sqrt{3^{2}+2^{2}+3(3)+2(2)+1}=3 \\sqrt{3}\\)
\nThus the required circle is
\n(x – 3)2<\/sup> + (y – 2)2<\/sup> = (3\\(\\sqrt{3}\\))2<\/sup>
\nx2<\/sup> + y2<\/sup> – 6x – 4y – 14 = 0.<\/p>\n","protected":false},"excerpt":{"rendered":"

Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B System of Circles Important Questions which are most likely to be asked in the exam. Intermediate 2nd Year Maths 2B System of Circles Important Questions Question 1. x2 + y2 + 4x + 8 = 0, x2 + y2 – 16y + k … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[14],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/20959"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=20959"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/20959\/revisions"}],"predecessor-version":[{"id":21042,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/20959\/revisions\/21042"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=20959"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=20959"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=20959"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}