Question

Suppose a circle passes through $\left(2,2\right)$ and $\left(9,9\right)$ and touches the $x-axisatP.$ If $O$ is the origin, then $OP$ is equal to

• A. $4$
• B. $5$
• C. $6$
• D. $9$
• E. $11$

Answer 1

The correct option is C

$6$

Explanation for the correct option:

Let the centre $C$ be $\left(h,k\right)$. Then $P=\left(h,0\right)$.

Here the radius $CP=r=\sqrt{{\left(h-h\right)}^2+{\left(k-0\right)}^2}$

$\Rightarrow CP=r=k$

Also $CP=CA=CB$

$\Rightarrow C{A}^2=C{B}^2$

$\Rightarrow {\left(h-2\right)}^2+{\left(k-2\right)}^2={\left(h-9\right)}^2+{\left(k-9\right)}^2$

$\Rightarrow h^2-4h+4+k^2-4k+4=h^2-18h+81+k^2-18k+81$

$\Rightarrow 14h+14k-154=0$

$\Rightarrow h+k=11$

Also $C{A}^2=r^2=k^2$

$\Rightarrow {\left(h-2\right)}^2+{\left(k-2\right)}^2=k^2$

$\Rightarrow h^2-4h+4+k^2-4k+4=k^2$

$\Rightarrow h^2-4\left(h+k\right)+8=0$

$\Rightarrow h^2-44+8=0$

$$\begin{gathered} \Rightarrow h^2=36 \\ \Rightarrow h=\pm 6 \end{gathered}$$

Since Circle is above $x-$ axis $h=6$

$\Rightarrow k=11-6=5$

Hence $OP=h=6$

Therefore option(C) is the correct answer