Question

If the straight line $y=x+4$ cuts the circle $x^2+y^2=26$ at $P$ and $Q$, where $Q$ is in first quadrant, then which of the following is/are correct?

Answer 1

Given $y=x+4$ cuts the circle $x^2+y^2=26$ at $P$ and $Q$
Finding the points $P$ and $Q$, we get
$x^2+(x+4{)}^2=26\Rightarrow 2x^2+8x-10=0\Rightarrow (x-1\left)\right(x+5)=0\Rightarrow x=-5,1x=-5\Rightarrow y=-1x=1\Rightarrow y=5$

So, the points are $P(-5,-1)$ and $Q(1,5)$

Mid point of $PQ$
$=(\frac{1-5}{2},\frac{5-1}{2})=(-2,2)$

Equation of tangent to circle $x^2+y^2=26$ is
$x{x}_1+y{y}_1-26=0$
Tangent at $P(-5,-1)$
$-5x-y-26=0\Rightarrow 5x+y+26=0$

Tangent at $Q(1,5)$
$x+5y-26=0$

The point of intersection of the tangents $y+5x+26=0$ and $5y+x-26=0$ is
$(-\frac{13}{2},\frac{13}{2})$.