 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 $2{{x}^{2}}+2{{y}^{2}}+4x-7y-25=0$ is.(a) $4{{x}^{2}}+4{{y}^{2}}+30x-13y-25=0$(b) $4{{x}^{2}}+4{{y}^{2}}+30x-13y+25=0$(c) $4{{x}^{2}}-4{{y}^{2}}-30x+13y-25=0$(d) $4{{x}^{2}}-4{{y}^{2}}+30x-13y-25=0$

Hint: In this question, we first need to write the general form for any circle passing through the point of intersection of other two circles from the family of circles . Now, by using the formula and substituting the given point we can get the assumed constant term in the equation from which we can get the equation of the circle.

CIRCLE: Circle is defined as the locus of a point which moves in a plane such that its distance from a fixed point in that plane is constant.
FAMILY OF CIRCLES: Any circle passing through the point of intersection of two circles S1 and S2 is
$\Rightarrow {{S}_{1}}+k{{S}_{2}}=0$
Now, by substituting the given two equations of the circles in the above formula we get,
$\Rightarrow {{x}^{2}}+{{y}^{2}}+13x-3y+k\left( 2{{x}^{2}}+2{{y}^{2}}+4x-7y-25 \right)=0$
Now, this will be the equation of the circle we need.
As given in the question that this circle passes through the point (1, 1).
Now, on substituting the values $x=1,y=1$in the above equation we get,
$\Rightarrow {{1}^{2}}+{{1}^{2}}+13\left( 1 \right)-3\left( 1 \right)+k\left( 2\left( {{1}^{2}} \right)+2\left( {{1}^{2}} \right)+4\left( 1 \right)-7\left( 1 \right)-25 \right)=0$
Now, from this on further simplification we get,
$\Rightarrow 1+1+13-3+k\left( 2+2+4-7-25 \right)=0$
Now, on adding and subtracting respectively we get,
$\Rightarrow 12-24k=0$
Now, on rearranging the terms we get,
$\Rightarrow 12=24k$
Let us now divide with 12 on both the sides.
$\therefore k=\dfrac{1}{2}$
Now, let us substitute this value of k in the general form to get the equation of the circle.
$\Rightarrow {{x}^{2}}+{{y}^{2}}+13x-3y+k\left( 2{{x}^{2}}+2{{y}^{2}}+4x-7y-25 \right)=0$
Let us now substitute the value of k in this equation.
$\Rightarrow {{x}^{2}}+{{y}^{2}}+13x-3y+\dfrac{1}{2}\left( 2{{x}^{2}}+2{{y}^{2}}+4x-7y-25 \right)=0$
Now, on multiplying the whole equation with 2 we get,
$\Rightarrow 2{{x}^{2}}+2{{y}^{2}}+26x-6y+2{{x}^{2}}+2{{y}^{2}}+4x-7y-25=0$
Now, on further simplifying this equation we get,
$\Rightarrow 4{{x}^{2}}+4{{y}^{2}}+30x-13y-25=0$
Hence, the correct option is (a).

Note: Instead of using the general form for the circle passing through the intersection of two circles from the family of circles we can also solve for the points of intersection and then find the equation of the circle passing through three points. But, this process will be a little complicated.
It is important to note that while substituting the given point in the equation and while finding the value of k we should be avoiding any calculation mistakes and should not neglect any of the terms because it changes the value of k and further the equation of the circle also.