Question

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

• A. $4x^2-4y^2-30x+13y-25=0$
• B. $4x^2-4y^2+30x-13y-25=0$
• C. $4x^2+4y^2+30x-13y+25=0$
• D. $4x^2+4y^2+30x-13y-25=0$

Answer 1

Answer: (D)

The correct option is A $4x^2+4y^2+30x-13y-25=0$
the equation of any curve through the points of intersection of the circles (1) and (2) will be

$(x^2+y^2+13x-3y)+k(2x^2+2y^2+4x-7y-25)=0$

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

$(1+1+13-3)+k(2+2+4-7-25)=0$

$12-24k=0$

$k=\frac{1}{2}$

Now, Subsitute the value of k in an first equation, we get

$(x^2+y^2+13x-3y)+\frac{1}{2}(2x^2+2y^2+4x-7y-25)=0$

$2x^2+2y^2+15x-\frac{13y}{2}-\frac{25}{2}=0$

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