Suppose a circle passes through $(2,2)$ and $(9,9)$ and touches the $X$-axis at $P$. If $O$ is the origin, then $O P$ is equal to

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Aicharya · Accepted Answer

Since, the circle passes through $A(2,2)$ and $B(9,9)$ and it touches on $X$-axis at $P$.

$\therefore \quad O P^2=O A \cdot O B$ ...(i) $\begin{aligned} & O A=\sqrt{2^2+2^2}=2 \sqrt{2} \\ & O B=\sqrt{9^2+9^2}=9 \sqrt{2}\end{aligned}$ $\begin{array}{ll}\therefore & O P^2=2 \sqrt{2} \cdot 9 \sqrt{2}=36 \\ \therefore & O P=6\end{array}$

Suppose a circle passes through $(2,2)$ and $(9,9)$ and touches the $X$-axis at $P$. If $O$ is the origin,…

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