Question

$PR$ and $PQ$ are tangents from $P$ to the center with centre $O$. If $BC$ is tangent to the circle at $X$. Prove that $PC+CX=PB+BX$.

Answer 1

R.E.F image

$PR=PQ$

So, $PC+CR$

$PB+BQ$....(1)

As tangent from extant point of circle are equal

So, : $CX=CR$

and $xb=bq$....(2)

So $PC+CR=PB+BQ$ (from (1))

$PC+CX=PB+XB$ (from (2))

Hence proved

$e{q}^n$ of circle : $x^2+y^2-6x-10y+\lambda$

radius of circle $(-g,f)\Rightarrow (3,5)$

$\therefore$ radius circle 73

$\therefore \lambda 752,\lambda 725$