Question

The equation of a circle which touches the lines $x=y$ at origin and passes through the point $\left(2,1\right)$ is $x^2+y^2+px+qy=0$ Then the value of $p,q$ are

• A. $5,5$
• B. $-5,5$
• C. $5,-5$
• D. None of these

Answer 1

The correct option is B

$-5,5$

Explanation of correct answer :

Finding values of p, q :

Let the equation of the circle passing through origin be $x^2+y^2+2gx+2fy=0$

It also passes through $\left(2,1\right)$.

$$\begin{gathered} \Rightarrow 4+1+4g+2f=0 \\ \Rightarrow 4g+2f=-5\dots \left(1\right) \end{gathered}$$

Also, the circle touches the line, $x=y$.

Perpendicular from centre $\left(-g,-f\right)$ to the tangent = Radius

$\Rightarrow \frac{\left|-f+g\right|}{\sqrt{1^2+1^2}}=\sqrt{g^2+f^2}$

Squaring both sides, we get

$$\begin{gathered} \Rightarrow f^2+g^2-2fg=2\left(g^2+f^2\right) \\ \Rightarrow f^2+g^2-2fg=2g^2+2f^2 \\ \Rightarrow f^2+g^2+2fg=0 \\ \Rightarrow {\left(f+g\right)}^2=0 \\ \Rightarrow f+g=0 \\ \Rightarrow g=-f \end{gathered}$$

From eqn $\left(1\right)$,

$$\begin{gathered} \Rightarrow 4\left(-f\right)+2f=-5 \\ \Rightarrow -2f=-5 \\ \Rightarrow f=\frac{5}{2} \\ \Rightarrow g=\frac{-5}{2}\left(\because g=-f\right) \end{gathered}$$

So, the equation of the circle is $x^2+y^2-5x+5y=0$

On comparing with,$x^2+y^2+px+qy=0$, we get

Thus, $p=-5,q=5$.

Hence, the correct answer is Option(B).