The equation of the circle which passes through the point of intersection of the circles x2+y2-8 x-2 y+7=0 and x2+y2-4 x+10 y+8=0 and having its centre on y-axis will be

Question

The equation of the circle which passes through the point of intersection of the circles x2+y2-8 x-2 y+7=0 and x2+y2-4 x+10 y+8=0 and having its centre on y-axis will be (A) x2+y2+22 x+9=0 (B) x2+y2+22 x-9=0 (C) x2+y2+22 y+9=0 (D) x2+y2+22 y-9=0

Tardigrade Admin · Accepted Answer

Equation of circles passing through point of intersection of given two circles is

x^{2} + y^{2} - 8 x - 2 y + 7 + λ (x^{2} + y^{2} - 4 x + 10 y + 8) = 0

... (i)
Centre of circle is

(4 + 2 λ, 1 - 5 λ)

.
Given that the centre lies on the

y

-axis then

4 + 2 λ = 0

\Rightarrow λ = - 2

Hence, the required equation is

x^{2} + y^{2} + 22 y + 9 = 0

.

Q. The equation of the circle which passes through the point of intersection of the circles $x^{2} + y^{2} - 8 x - 2 y + 7 = 0$ and $x^{2} + y^{2} - 4 x + 10 y + 8 = 0$ and having its centre on $y$ -axis will be

$x^{2} + y^{2} + 22 x + 9 = 0$

$x^{2} + y^{2} + 22 x - 9 = 0$

$x^{2} + y^{2} + 22 y + 9 = 0$

$x^{2} + y^{2} + 22 y - 9 = 0$

Q. The equation of the circle which passes through the point of intersection of the circles x2+y2−8x−2y+7=0 and x2+y2−4x+10y+8=0 and having its centre on y-axis will be

x2+y2+22x+9=0

x2+y2+22x−9=0

x2+y2+22y+9=0

x2+y2+22y−9=0

Equation of circles passing through point of intersection of given two circles is x2+y2−8x−2y+7+λ(x2+y2−4x+10y+8)=0 ... (i) Centre of circle is (4+2λ,1−5λ). Given that the centre lies on the y-axis then 4+2λ=0 ⇒λ=−2 Hence, the required equation is x2+y2+22y+9=0.

Q. The equation of the circle which passes through the point of intersection of the circles $x^{2} + y^{2} - 8 x - 2 y + 7 = 0$ and $x^{2} + y^{2} - 4 x + 10 y + 8 = 0$ and having its centre on $y$ -axis will be

$x^{2} + y^{2} + 22 x + 9 = 0$

$x^{2} + y^{2} + 22 x - 9 = 0$

$x^{2} + y^{2} + 22 y + 9 = 0$

$x^{2} + y^{2} + 22 y - 9 = 0$