Question

Point on the straight line $x-y+1=0$, the tangents from which to the circle $x^2+y^2-3x=0$ are of $2$ unit is

• A. $(\frac{-3}{2},\frac{5}{2})$
• B. $(\frac{-3}{2},\frac{-5}{2})$
• C. $(\frac{3}{2},\frac{5}{2})$
• D. $(\frac{3}{2},\frac{-5}{2})$

Answer 1

Answer: (A)

$(\frac{-3}{2},\frac{5}{2})$

Let the point on the line be $P(h,k)$

$h+1=k\left(1\right)$
Length of tangent $=2=\sqrt{S_{11}}$
$\sqrt{h^2+k^2-3h}=2h^2+k^2-3h=4\left(2\right)$

Puttong value of $k$ from $\left(1\right)$ in $\left(2\right)$

$h^2+(h+1{)}^2-3h=42h^2+2h-3h-3=02h(h+1)-3(h+1)=0(h+1\left)\right(2h-3)=0h=-1,\frac{3}{2}$

gives $k=0,\frac{5}{2}$

Coordinates of point $P$ are $(-1,0)$ and $(\frac{3}{2},\frac{5}{2})$