Suppose a circle passes through (2,2) and (9, | vedclass

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(C) Let the circle pass through points $A(2,2)$ and $B(9,9)$ and touch the $X$-axis at point $P$. By the power of a point theorem,since $OP$ is a tang

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$6$

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(C) Let the circle pass through points $A(2,2)$ and $B(9,9)$ and touch the $X$-axis at point $P$. By the power of a point theorem,since $OP$ is a tangent to the circle from the origin $O$ and $OAB$ is a secant line,we have: $OP^2 = OA \cdot OB$ Calculating the distances $OA$ and $OB$ from the origin $O(0,0)$: $OA = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}$ $OB = \sqrt{9^2 + 9^2} = \sqrt{81 + 81} = \sqrt{162} = 9\sqrt{2}$ Substituting these values into the equation: $OP^2 = (2\sqrt{2}) \cdot (9\sqrt{2}) = 18 \cdot 2 = 36$ $OP = \sqrt{36} = 6$ Thus,$OP = 6$.

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

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