Consider the circle $x^2+y^2-4 x-2 y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the line segment $P A$ meets the circle in $Q$ with $P Q=5$, then $c$ is equal to

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Aicharya · Accepted Answer

Given equation of circle is $x^2+y^2-4 x-2 y+c=0$ whose centre is $A(2,1)$.

Now, $A P=\sqrt{(2-10)^2+(1-7)^2}$ $\begin{aligned} & =\sqrt{(-8)^2+(-6)^2}=\sqrt{64+36}=\sqrt{100} \\ & =10\end{aligned}$ $\therefore \quad A Q=A P-P Q=10-5=5$ So, $Q$ is the mid-point of AP $=\left(\frac{10+2}{2}, \frac{7+1}{2}\right)=(6,4)$ Since, $Q$ lies on a circle. $\begin{array}{rlrl} & \therefore & 6^2+4^2-4(6)-2(4)+c & =0 \\ & \Rightarrow & 36+16-24-8+c & =0 \\ & \Rightarrow & 20+c & =0 \\ \Rightarrow & c & =-20\end{array}$

Consider the circle $x^2+y^2-4 x-2 y+c=0$ whose centre is $A(2,1)$. If the point $P(10,7)$ is such that the…

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