Question

In the given figure, the circle $x^2+y^2=25$ 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 $AQ:QP=5:3$

Answer 1

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

The equation $x=11$ Touches $\text{x-axis}$ 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

$x_2=\frac{5\left(11\right)+3(-5)}{5+3}{x}_2=\frac{55-15}{8}{x}_2=\frac{40}{8}=5$

$Q$ lies on Circle

$x_2^2+y_2^2=25⟹y_2=0$

Therefore $Q$ is $(5,0)$