In the given figure, the circle $x^{2} + y^{2} = 25$ intersects the $x - a x i s$ at the point $A$ and $B$ . The line $x = 11$ intersects the $x - a x i s$ at the point $C$ .Point $p$ moves along the $x = 11$ above the $x - a x i s$ and $A P$ intersects the circle at $Q$ .Find $Q$ when $A Q : Q P = 5 : 3$

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Solution

The equation $x^{2} + y^{2} = 25$ Touches $x-axis$ at $A (- 5, 0)$ and $B (5, 0)$

The equation $x = 11$ Touches $x-axis$ at $C (11, 0)$
Let $P$ be $(x_{1}, y_{1})$ and $Q$ be $(x_{2}, y_{2})$
Since $P$ moves all way on Line $x = 11 ⟹ x_{1} = 11$
Considering Section Formulae
$x_{2} = \frac{5 (11) + 3 (- 5)}{5 + 3} x_{2} = \frac{55 - 15}{8} x_{2} = \frac{40}{8} = 5$
$Q$ lies on Circle
$x_{2}^{2} + y_{2}^{2} = 25 ⟹ y_{2} = 0$
Therefore $Q$ is $(5, 0)$

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In the given figure, the circle x2+y2=25 intersects the x−axis at the point A and B. The line x=11 intersects the x−axis at the point C.Point p moves along the x=11 above the x−axis and AP intersects the circle at Q.Find Q when AQ:QP=5:3