The equation of a circle passing through the point $(1, 1)$ and the point of intersection of the circles
$x^{2} + y^{2} + 13 x - 3 y = 0$ and $2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0$ is

4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0

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4 x^{2} + 4 y^{2} + 30 x - 13 y + 25 = 0

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4 x^{2} - 4 y^{2} - 30 x + 13 y - 25 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 x^{2} - 4 y^{2} + 30 x - 13 y - 25 = 0

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Solution

The correct option is D $4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0$
the equation of any curve through the points of intersection of the circles (1) and (2) will be
$(x^{2} + y^{2} + 13 x - 3 y) + k (2 x^{2} + 2 y^{2} + 4 x - 7 y - 25) = 0$
It is given that equation of circles passes through the point $(1, 1)$
So,
$x = 1$ and $y = 1$
Substitute these value in an above equation, we get
$(1 + 1 + 13 - 3) + k (2 + 2 + 4 - 7 - 25) = 0$
$12 - 24 k = 0$
$k = \frac{1}{2}$
Now, Subsitute the value of k in an first equation, we get
$(x^{2} + y^{2} + 13 x - 3 y) + \frac{1}{2} (2 x^{2} + 2 y^{2} + 4 x - 7 y - 25) = 0$
$2 x^{2} + 2 y^{2} + 15 x - \frac{13 y}{2} - \frac{25}{2} = 0$
$4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0$

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Similar questions

x^{2} + y^{2} + 13 x - 3 y = 0

2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0

The equation of a circle passing through points of intersection of the circles $x^{2} + y^{2} + 13 x - 3 y = 0$ and

$2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0$ and point (1, 1) is

(1, 1)

x^{2} + y^{2} + 13 x - 13 y = 0

2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0

4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0

Find the equation of pair of tangents drawn to the circle $x^{2} + y^{2} - 4 x + 4 y = 0$ from the point (2, 2)

(1, 1)

x^{2} + y^{2} + 13 x - 3 y = 0

2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0

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