Inter 1st Year Maths Conic Sections Solutions Exercise 10a

Practicing the AP Board Solutions Class 11 Maths and Chapter 10 Inter 1st Year Maths Conic Sections Solutions Exercise 10a Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Conic Sections Solutions Exercise 10a

Conic Sections Exercise 10a Solutions

Conic Sections Class 11 Exercise 10a Solutions – Conic Sections 10a Exercise Solutions

I. In each of the following exercises 1 to 5, find the equation of the circle with

Question 1.
Centre (0, 2) and radius 2.
Solution:
Equation of the circle is (x – 0)² + (y – 2)² = 2²
⇒ x² + y² + 4 – 4y = 4
⇒ x² + y² – 4y = 0

Question 2.
Centre (-2, 3) and radius 4.
Solution:
Equation of the circle is (x + 3)² + (y – 2)² = 4²
⇒ x² + 6x + 9 + y² – 4y + 4 = 16
⇒ x² + y² + 6x – 4y – 3 = 0

Question 3.
Centre \(\left(\frac{1}{2}, \frac{1}{4}\right)\) and radius \(\frac {1}{12}\).
Solution:
The equation of the circle is

⇒ 9(16x² + 16y² – 16x – 8y + 5) = 1
⇒ 144(x² + y²) – 144x – 72y + 44 = 0
⇒ 36x² + 36y² – 36x – 18y + 11 = 0

Question 4.
Centre (1, 1) and radius √2.
Solution:
Equation of the circle is (x – 1)² + (y – 1)² = (√2)²
⇒ x² – 2x + 1 + y² – 2y + 1 = 2
⇒ x² + y² – 2x – 2y = 0

Question 5.
Centre (-a, -b) and radius \(\sqrt{a^2-b^2}\).
Solution:
The equation of the circle is (x – (-a))² + (y – (-b))² = \(\left(\sqrt{a^2-b^2}\right)^2\)
⇒ x² + 2ax + a² + y² + 2by + b² = a² – b²
⇒ x² + y² + 2ax + 2by + 2b² = 0

In each of the following Exercises 6 to 10, find the centre and radius of the circles.

Question 6.
(x + 5)² + (y – 3)² = 36
Solution:
Equation of the given circle is (x + 5)² + (y – 3)² = 36
It is of the form (x – h)² + (y – k)² = r²
Here h = -5, k = 3, and r = 6.
Thus, the centre of the given circle is (-5, 3) and the radius is 6.

Question 7.
x² + y² – 4x – 8y – 45 = 0
Solution:
Equation of the given circle is x² + y² – 4x – 8y – 45 = 0
⇒ (x² – 4x) + (y² – 8y) = 45
⇒ {x² – 2(x)(2) + 2²} + {y² – 2(y)(4) + (4)²} – 4 – 16 = 45
⇒ (x – 2)² + (y – 4)² = (√65)²
which is of the form (x – h)² + (y – k)² = r²
where h = 2, k = 4, and r = √65
Thus, the centre of the given circle is (2, 4) and the radius is √65.

Question 8.
x² + y² – 8x + 10y – 12 = 0
Solution:
Equation of the given circle is x² + y² – 8x + 10y – 12 = 0
⇒ (x² – 8x) + (y² + 10y) = 12
⇒ {x² – 2(x)(4) + 4²} + {y² + 2(y)(5) + 5²} – 16 – 25 = 12
⇒ (x – 4)² + (y + 5)² = 53
⇒ (x – 4)² + (y – (-5))² = (√53)²
which is of the form (x – h)² + (y – k)² = r²
where h = 4, k = -5 and r = √53
Thus, the centre of the given circle is (4, -5) and the radius is √53.

Question 9.
2x² + 2y² – x = 0
Solution:
Equation of the given circle is 2x² + 2y² – x = 0
⇒ (2x² – x) + 2y² = 0

which is of the form (x – h)² + (y – k)² = r²
Here h = \(\frac {1}{4}\), k = 0, and r = \(\frac {1}{4}\)
Thus, the centre of the given circle is (\(\frac {1}{4}\), 0) and radius is \(\frac {1}{4}\).

Question 10.
\(\sqrt{1+\mathrm{m}^2}\)(x² + y²) – 2cx – 2mcy = 0
Solution:
\(\sqrt{1+\mathrm{m}^2}\)(x² + y²) – 2cx – 2mcy = 0
Comparing the given equation with Ax² + Ay² + 2Gx + 2Fy + C = 0, we get
A = \(\sqrt{1+\mathrm{m}^2}\), G = -c, F = -cm, C = 0
Centre = \(\left(\frac{-G}{A}, \frac{-F}{A}\right)=\left(\frac{c}{\sqrt{1+m^2}}, \frac{c m}{\sqrt{1+m^2}}\right)\)

Question 11.
Find the equation of the circle passing through the origin and having the centre at (-4, -3).
Solution:
Let C(-4, -3) and P(0, 0)
Radius = CP
= \(\sqrt{(-4)^2+(-3)^2}\)
= 5
Required circle equation is (x + 4)² + (y + 3)² = 5²
⇒ x² + 8x + 16 + y² + 6y + 9 = 25
⇒ x² + y² + 8x + 6y = 0

Question 12.
Find the equation of the circle passing through (2, -1) having the centre at (2, 3).
Solution:
Let C(2, 3) and P(2, -1)
Radius = CP
= \(\sqrt{(2-2)^2+(3+1)^2}\)
= 4
Required circle equation is (x – 2)² + (y – 3)² = 4²
⇒ x² – 4x + 4 + y² – 6y + 9 = 16
⇒ x² + y² – 4x – 6y – 3 = 0

Question 13.
Find the equation of the circle passing through (-2, 3) having the centre at (0, 0).
Solution:
Let C(0, 0) and P(-2, 3)
Radius = CP
= \(\sqrt{(-2)^2+3^2}\)
= √13
Required circle equation is x² + y² = (√13)²
⇒ x² + y² = 13

Question 14.
Find the equation of the circle passing through (3, 4) and having the centre at (-3, 4).
Solution:
Centre of the circle (-3, 4)
The equation of a circle with centre (b, k) and radius r is (x – b)² + (y – k)² = r²
⇒ (x + 3)² + (y – 4)² = r²
r = \(\sqrt{\left(3-(-3)^2+(4-4)^2\right.}\)
= \(\sqrt{6^2+0}\)
= 6
⇒ r² = 36
∴ Equation of the circle is (x + 3)² + (y – 4)² = 36.

Question 15.
Find the value of a if 2x² + ay² – 3x + 2y – 1 = 0 represents a circle, and also find its radius.
Solution:
2x² + ay² – 3x + 2y – 1 = 0
The equation represents a circle
coefficient of x² = coefficient of y²
⇒ a = 2
Comparing the given equation with ax² + ay² + 2gx + 2fy + c = 0, we get

Question 16.
Find the values of a and b if ax² + bxy + 3y² – 5x + 2y – 3 = 0 represents a circle. Also, find the radius and centre of the circle.
Solution:
The equation represents a circle
coefficient of x² = coefficient of y²
and coefficient of xy = 0
⇒ a = 3, b = 0
Comparing 3x² + 3y² – 5x + 2y – 3 = 0 with ax² + ay² + 2gx + 2fy + c = 0
we get a = 3, g = \(\frac {-5}{2}\), f = 1, c = -3

Question 17.
If x² + y² + 2gx + 2fy – 12 = 0 represents a circle with centre (2, 3), find g, f, and its radius.
Solution:
Centre of x² + y² + 2gx + 2fy – 12 = 0 is (2, 3)
⇒ (-g, -f) = (2, 3)
⇒ g = -2, f = -3

Question 18.
If x² + y² + 2gx + 2fy = 0 represents a circle with centre (-4, -3), then find g, f, and the radius of the circle.
Solution:
Centre of x² + y² + 2gx + 2fy = 0 is (-4, -3)
⇒ (-g, -f) = (-4, -3)
⇒ g = 4, f = 3
Radius = \(\sqrt{g^2+\mathrm{f}^2-\mathrm{c}}\)
= \(\sqrt{4^2+3^2-0}\)
= \(\sqrt{16+9}\)
= 5

Question 19.
If x² + y² – 4x + 6y + c = 0 represents a circle with radius 6, then find the value of c.
Solution:
Radius of x² + y² – 4x + 6y + c = 0 is 6
⇒ \(\sqrt{(-2)^2+(3)^2-c}\) = 6
⇒ 4 + 9 – c = 36
⇒ c = -23

Question 20.
Find the equation of a circle that is concentric with x² + y² – 6x – 4y – 12 = 0 and passing through (-2, 14).
Solution:
Equation of the circle concentric with x² + y² – 6x – 4y -12 = 0 is S = x² + y² – 6x – 4y + k = 0
This circle passes through (-2, 14)
⇒ S(-2, 14) = 0
⇒ (-2)² + (14)² – 6(-2) – 4(14) + k = 0
⇒ 4 + 196 + 12 – 56 + k = 0
⇒ k = -156
∴ The required circle is x² + y² – 6x – 4y – 156 = 0.

Question 21.
Find the equation of a circle with centre (2, 2) and passing through the point (4, 5).
Solution:
The centre of the circle is (h, k) = (2, 2)
Since the circle passes through point (4, 5), then the radius (r) of the circle is the distance between the points (2, 2) and (4, 5)
∴ r = \(\sqrt{(2-4)^2+(2-5)^2}\)
= \(\sqrt{(-2)^2+(-3)^2}\)
= \(\sqrt{4+9}\)
= √13
Thus, the equation of the circle is (x – h)² + (y – k)² = r²
⇒ (x – 2)² + (y – 2)² = (√13)²
⇒ x² – 4x + 4 + y² – 4y + 4 = 13
⇒ x² + y² – 4x – 4y – 5 = 0

Question 22.
Does the point (-2.5, 3.5) lie inside, outside, or on the circle x² + y² = 25?
Solution:
The equation of the circle is x² + y² = 25
Centre = (0, 0) and radius = 5
Distance between point (-2.5, 3.5) and centre (0, 0) is \(\sqrt{(-2.5-0)^2+(3.5-0)^2}\)
= \(\sqrt{6.25+12.25}\)
= \(\sqrt{18.5}\)
= 4.3 (approx)
< 5 = r
Since the distance between point (-2.5, 3.5) and the centre (0, 0) of the circle is less than the radius, the point (-2.5, 3.5) lies inside the circle.

II.

Question 1.
If the abscissae of points A, B are the roots of the equation x² + 2ax – b² = 0, and the ordinates of A, B are roots of y² + 2py – q² = 0, then find the equation of a circle for which \(\overline{\mathrm{AB}}\) is a diameter.
Solution:
Let A(x₁, y₁), B(x₂, y₂)
x₁, x₂ are the roots of x² + 2ax – b² = 0
⇒ x₁ + x₂ = -2a, x₁x₂ = -b²
y₁, y₂ are the roots of y² + 2py – q² = 0
⇒ y₁ + y₂ = -2p, y₁y₂ = -q²
The equation of the circle with AB as diameter is (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0
⇒ x² + y² – (x₁ + x₂)x – (y₁ + y₂)y + (x₁x₂ + y₁y₂) = 0
⇒ x² + y² – (-2a)x – (-2p)y + (-b² – q²) = 0
⇒ x² + y² + 2ax + 2py – (b² + q²) = 0

Question 2.
Show that A(3, -1) lies on the circle x² + y² – 2x + 4y = 0. Also, find the other end of the diameter through A.
Solution:
Let S = x² + y² – 2x + 4y = 0
S(3, -1) = (3)² + (-1)² – 2(3) + 4(-1)
= 9 + 1 – 6 – 4
= 0
S(3, -1) = 0
⇒ (3, -1) lies on S = 0.
Cenure or me circle, C = (1, -2)
Let A(3, -1), B(α, β) be the ends of a diameter.
∴ The midpoint of AB is C
⇒ \(\left(\frac{3+\alpha}{2}, \frac{-1+\beta}{2}\right)\) = (1, -2)
⇒ \(\frac{3+\alpha}{2}\) = 1, \(\frac{-1+\beta}{2}\) = -2
⇒ 3 + α = 2, -1 + β = -4
⇒ α = -1, β = -3
∴ The other end is (-1, -3).

Question 3.
Show that A(-3, 0) lies on x² + y² + 8x + 12y + 15 = 0 and find the other end of the diameter through A.
Solution:
Let S = x² + y² + 8x + 12y + 15 = 0
S(-3, 0)
= (-3)² + 0 + 8(-3) + 12(0) + 15
= 9 – 24 + 15
= 0
S(-3, 0) = 0
⇒ (-3, 0) lies on S = 0
Centre of the circle is C(-4, -6)
Let A(-3, 0), B(α, β) be the ends of a diameter.
∴ The midpoint of AB is C
⇒ \(\left(\frac{-3+\alpha}{2}, \frac{0+\beta}{2}\right)\) = (-4, -6)
⇒ \(\frac{\alpha-3}{2}\) = -4, \(\frac{\beta}{2}\) = -6
⇒ α – 3 = -8, β = -12
⇒ α = -5, β = -12
∴ The other end is (-5, -12)

Question 4.
Find the equation of the circle whose centre lies on the X-axis and passes through (-2, 3) and (4, 5).
Solution:
Let the required circle be
S = x² + y² + 2gx + 2fy + c = 0
(-2, 3) lies on S = 0
⇒ (-2)² + 3² + 2g(-2) + 2f(3) + c = 0
⇒ 4g + 6f + c = -13 ………..(1)
(4, 5) lies on S = 0
⇒ 4² + 5² + 2g(4) + 2f(5) + c = 0
⇒ 8g + 10f + c = -41 ……..(2)
(2) – (1) ⇒ 12g + 4f = -28
⇒ 3g + f = -7 …….(3)
Centre (-g, -f) of S = 0 lies on x-axis
⇒ f = 0
From (3), 3g = -7
⇒ g = \(\frac {-7}{3}\)
From (1), c = -13 + 4g – 6f
= -13 + \(4\left(\frac{-7}{3}\right)\) – 6(0)
= \(\frac{-39-28}{3}\)
= \(\frac {-67}{3}\)
∴ The equation of the required circle is x² + y² + \(2\left(\frac{-7}{3}\right) \times-\frac{67}{3}\) = 0
⇒ 3(x² + y²) – 14x – 67 = 0

III.

Question 1.
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
Solution:
Let the required circle be
S = x² + y² + 2gx + 2fy + c = 0
(4, 1) lies on S = 0
⇒ (4)² + (1)² + 2g(4) + 2f(1) + c = 0
⇒ 16 + 1 + 8g + 2f + c = 0
⇒ 8g + 2f + c = -17 ………(1)
(6, 5) lies on S = 0
⇒ (6)² + (5)² + 2g(6) + 2f(5) + c = 0
⇒ 36 + 25 + 12g + 10f + c = 0
⇒ 12g + 10f + c = -61 ……….(2)
(2) – (1) ⇒ -4g – 8f = 44 ……….(3)
Centre (-g, -f) of S = 0 lies on 4x + y = 16
⇒ 4(-g) + (-f) = 16
⇒ -4g – f = 16 …….. (4)
Solving (3) & (4):
-7f = 28
⇒ f = -4
(4) ⇒ 4g = -f – 16
⇒ 4g = -(-4) – 16
⇒ 4g = -12
⇒ g = -3
(1) ⇒ 8g + 2f + c = -17
⇒ g(-3) + 2(-4) + c = 17
⇒ -24 – 8 + c = 17
⇒ -32 + c = 17
⇒ c = 7 + 32
⇒ c = 49
∴ The equation of required circle is x² + y² + 2(-3)x + 2(-4)y + 49 = 0
⇒ x² + y² – 6x – 8y + 49 = 0.

Question 2.
Find the equation of the circle passing through the points (2, 3) and (-1, 1) and whose centre is on the line x – 3y – 11 = 0.
Solution:
Let the equation of the circle be (x – h)² + (y – k)² = r²
The circle passes through points (2, 3)
⇒ (2 – h)² + (3 – k)² = r²
⇒ h² + k² – 4h – 6k + 13 = r² ……….(1)
The circle passes through points
⇒ (-1 – h)² + (1 – k)² = r²
⇒ h² + k² + 2h – 2k + 2 = r² ……….(2)
The centre (h, k) lies on the line x – 3y – 11 = 0
⇒ h – 3k = 11 ………(3)
(1) – (2) ⇒ 6h + 4k = 11 ………(4)
On solving equations (3) and (4), we obtain
h = \(\frac {7}{2}\) and k = \(\frac {-5}{2}\)
On substituting the values of h and k in equation (1), we obtain

⇒ 4x² – 28x + 49 + 4y² + 20y + 25 = 130
⇒ 4x² + 4y² – 28x + 20y – 56 = 0
⇒ 4(x² + y² – 7x + 5y – 14) = 0
⇒ x² + y² – 7x + 5y – 14 = 0

Question 3.
Find the equation of a circle that passes through (2, -3) and (-4, 5) and has the centre on 4x + 3y + 1 = 0.
Solution:
Let the required circle be
S = x² + y² + 2gx + 2fy + c = 0
(2, -3) lies on S = 0
⇒ 2² + (-3)² + 2g(2) + 2f(-3) + c = 0
⇒ 4g – 6f + c = -13 ……..(1)
(-4, 5) lies on S = 0
⇒ (-4)² + 5² + 2g (-4) + 2f(5) + c = 0
⇒ -8g + 10f + c = -41 ……..(2)
(1) – (2) ⇒ 12g – 16f = 28
⇒ 3g – 4f = 7 ……(3)
Centre (-g, -f) of S = 0 lies on 4x + 3y + 1 = 0
⇒ 4(-g) + 3(-f) + 1 = 0
⇒ 4g + 3f = 1 …….. (4)
Solving (3) & (4);
g = 1, f = -1
From (1), c = -13 – 4g + 6f
= -13 – 4(1) + 6(-1)
= -23
∴ The equation of the required circle is x² + y² + 2x – 2y – 23 = 0.

Question 4.
Find the equation of a circle that passes through (4, 1) and (6, 5) and has the centre on 4x + 3y – 24 = 0.
Solution:
Let the required circle be S = x² + y² + 2gx + 2fy + c = 0
(4, 1) lies on S = 0
⇒ (4)² + (1)² + 2g(4) + 2f(1) + c = 0
⇒ 16 + 1 + 8g + 2f + c = 0
⇒ 8g + 2f + c = -17 ………(1)
(6, 5) lies on S = 0
⇒ (6)² + (5)² + 2g(6) + 2f(5) + c = 0
⇒ 36 + 25 + 12g + 10f + c = 0
⇒ 12g + 10f + c = -61 ………(2)
(2) – (1) ⇒ -4g – 8f = 44 ……..(3)
Centre (-g, -f) of S = 0 lies on 4x + 3y – 24 = 0
⇒ 4(-g) + 3(-f) = 24
⇒ -4g – 3f = 24 …… (4)
Solving (3) & (4)
⇒ -5f = 20
⇒ f = -4
(4) ⇒ -4g – 3(-4) = 24
⇒ -4g + 12 = 24
⇒ -4g = 24 – 12
⇒ -4g = 12
⇒ g = -3
(1) ⇒ 8g + 2f + c = -17
⇒ 8(-3) + 2(-4) + c = -17
⇒ -24 – 8 + c = -17
⇒ -32 + c = -17
⇒ c = -17 + 32
⇒ c = 15
∴ The equation of required circle is x² + y² + 2(-3)x + 2(-4)y + 15 = 0
⇒ x² + y² – 6x – 4y + 15 = 0

Question 5.
Find the equation of the circle with radius 5 whose centre lies on the x-axis and passes through the point (2, 3).
Solution:
Let the equation of the required circle be (x – h)² + (y – k)² = r²
Since the radius of the circle is 5 and its centre lies on the x-axis,
k = 0 and r = 5.
Now, the equation of the circle becomes (x – h)² + y² = 25.
It is given that the circle passes through the point (2, 3)
∴ (2 – h)² + 3² = 25
⇒ (2 – h)² = 25 – 9
⇒ (2 – h)² = 16
⇒ 2 – h = ±4
⇒ h = 6 or -2
When h = -2, the equation of the circle becomes (x + 2)² + y² = 25
⇒ x² + 4x + 4 + y² = 25
⇒ x² + y² + 4x – 21 = 0
When h = 6, the equation of the circle becomes (x – 6)² + y² = 25
⇒ x² – 12x + 36 + y² = 25
⇒ x² + y² – 12x + 11 = 0

Question 6.
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
Solution:
Let the equation of the circle be (x – h)² + (y – k)² = r²
The circle passes through (0, 0)
⇒ (0 – h)² + (0 – k)² = r²
⇒ h² + k² = r²
The equation of the circle now becomes (x – h)² + (y – k)² = h² + k²
It is given that the circle makes intercepts a and b on the coordinate axes.
This means that the circle passes through points (a, 0) and (0, b).
Therefore, (a – h)² + (0 – k)² = h² + k²
⇒ a² – 2ah = 0
⇒ h = \(\frac {a}{2}\)
and (0 – h)² + (b – k)² = h² + k²
⇒ b² – 2bk = 0
⇒ k = \(\frac {b}{2}\)
Thus, the equation of the circle is
⇒ \(\left(x-\frac{a}{2}\right)^2+\left(y-\frac{b}{2}\right)^2=\left(\frac{a}{2}\right)^2+\left(\frac{b}{2}\right)^2\)
⇒ x² + y² – ax – by + \(\frac{a^2}{4}+\frac{b^2}{4}=\frac{a^2}{4}+\frac{b^2}{4}\)
⇒ x² + y² – ax – by = 0

Question 7.
Find the equation of the circle passing through each of the following three points.
(i) (0, 0), (2, 0), (0, 2)
(ii) (3, 4), (3, 2), (1, 4)
(iii) (2, 1), (5, 5), (-6, 7)
Solution:
(i) Let O(0, 0), A(2, 0), B = (0, 2)
Let the required circle be S = x² + y² + 2gx + 2fy + c = 0
O lies on S = 0
⇒ S(0, 0) = 0
⇒ c = 0
A lies on S = 0
⇒ S (0, 0) = 0
⇒ 4 + 0 + 4g + 0 + c = 0
⇒ 4g + c = -4
⇒ 4g = -4
⇒ g = -1
B lies on S = 0
⇒ S(0, 2) = 0
⇒ 0 + 4 + 0 + 4f + c = 0
⇒ 4f + c = -4
⇒ 4f = -4
⇒ f = -1
∴ The equation to the required circle is x² + y² – 2x – 2y = 0

(ii) Let A(3, 4), B(3, 2), C(1, 4)
A lies on S = 0
⇒ S (2, 1) = 0
⇒ 4 + 1 + 4g + 2f + c = 0
⇒ 4g + 2f + c = -5 …….(1)
B lies on S = 0
⇒ S(5, 5) = 0
⇒ 25 + 25 + 10g + 10f + c = 0
⇒ 10g + 10f + c = -50 ……….(2)
C lies on S = 0
⇒ S (-6, 7) = 0
⇒ 36 + 49 – 12g + 14f + c = 0
⇒ -12g + 14f + c = -85 ………(3)
∴ The equation to the required circle is x² + y² – 6x – 6y + 17 + 2(x – 3) = 0
⇒ x² + y² – 4x – 6y + 11 = 0

(iii) We are given three points: A(2, 1), B(5, 5), C(-6, 7).
We are to find the equation of the circle passing through all three.
General form of a circle: x² + y² + Dx + Ey + F = 0
Substitute the three points into the general equation:
(1) Using (2, 1):
2² + 1² + 2D + E + F = 0
⇒ 4 + 1 + 2D + E + F = 0
⇒ 2D + E + F = -5 …….(1)
(2) Using (5, 5):
25 + 25 + 5D + 5E + F = 0
⇒ 5D + 5E + F = -50 ……….(2)
(3) Using (-6, 7):
36 + 49 – 6D + 7E + F = 0
⇒ -6D + 7E + F = -85 ………..(3)
Now, solve the system:
(2) – (1) ⇒ (5D + 5E + F) – (2D + E + F) = -50 + 5
⇒ 3D + 4E = -45 …….(4)
(3) – (1) ⇒ (-6D + 7E + F) – (2D + E + F) = -85 + 5
⇒ -8D + 6E = -80
⇒ -4D + 3E = -40 …….(5)
Now solve (4) and (5):
From (4) ⇒ 3D + 4E = – 45
From (5) ⇒ -4D + 3E = -40
Multiply (4) by 3 ⇒ 9D + 12E = -135
Multiply (5) by 4 ⇒ -16D + 12E = -160
Now subtract:
(9D + 12E) – (-16D + 12E) = -135 + 160
⇒ 25D = 25
⇒ D = 1
Now substitute D = 1 in (4):
3(1) + 4E = -45
⇒ 4E = -48
⇒ E = -12
Now substitute D = 1, E = -12 in (1):
2(1) – 12 + F = -5
⇒ F = 5
Final equation of the circle:
x² + y² + x – 12y + 5 = 0