Question

The circle passing through the intersection of the circles ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}=0$ and, $x^2+y^2-4y=0$, having its center on the line, $2\mathrm{x}-3\mathrm{y}+12=0,$ also passes through the point:

• A. $(-1,3)$
• B. $(1,-3)$
• C. $(-3,6)$
• D. $(-3,1)$

Answer 1

The correct option is C

$(-3,6)$

Explanation for the correct option:

Finding the equation of the circle:

Given that the equation of

circle-1: ${\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}=0$

circle-2: $x^2+y^2-4y=0$

Now, equation of circle passing through the intersection of the above circles, is given by

$\mathrm{S}1+\mathrm{\lambda }(\mathrm{S}1-\mathrm{S}2)=0$ [Using the concept of family of circles.]

$$\begin{gathered} {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}+\lambda (4y-6x)=0 \\ {\mathrm{x}}^2+{\mathrm{y}}^2-6\mathrm{x}(1+\lambda )+\left(4\lambda y\right)=0.....\left(i\right) \end{gathered}$$

Centre is $\left(3\right(1+\lambda ),-2\lambda )$

Since, centre lies on the line $2\mathrm{x}-3\mathrm{y}+12=0$, So

$$\begin{gathered} 6+6\mathrm{\lambda }+6\mathrm{\lambda }+12=0 \\ 12\mathrm{\lambda }=-18 \\ \lambda =\frac{-3}{2} \end{gathered}$$

Now putting the value of $\lambda$ in equation $\left(i\right)$, we get

The equation of the circle is ${\mathrm{x}}^2+{\mathrm{y}}^2+3\mathrm{x}-6\mathrm{y}=0$

Through the trial and error basis method, we have $(-3,6)$ is the point which lies on ${\mathrm{x}}^2+{\mathrm{y}}^2+3\mathrm{x}-6\mathrm{y}=0$

Therefore, the correct answer is option(C).