Question

Find the equation to the straight lines joining the origin to the points in which the straight $y=mx+c$ cuts the circle
$x^2+y^2=2ax+2by$
Hence find the condition that these points may subtend a right angle at the origin.
Find also the condition that the straight line may touch the circle.

Answer 1

PART 1)

Lets make a pair of straight lines $S^{\prime }$ of the lines joining origin to the points of intersection of the line $L$ and circle $S$.

$S:x^2+y^2-2ax-2by=0$

$L:y-mx-c=0$

The method for this is:

Step 1) Modify $L:\frac{y-mx}{c}=1$

Step 2) Multiply $1$ to terms with degree $1$ and $1^2$ to terms with degree $0$ ( Basically, we are making it into a homogeneous pair of straight lines)
So, $S^{\prime }:x^2+y^2-(2ax+2by)\left(\frac{y-mx}{c}\right)=0$

Solving, we get :

$S^{\prime }:(1+\frac{2am}{c})x^2+(1-\frac{2b}{c})y^2+\left(\frac{2bm-2a}{c}\right)xy$

The condition for the two lines to be perpendicular is : coefficient of $x^2$ + coefficient of $y^2$ = 0

$1+\frac{2am}{c}+1-\frac{2b}{c}=0$

Solving, we get:

$c=b-am$

PART2)

For the line to be a tangent, the perpendicular distance of the line from the center of the circle = radius of the circle.

Radius = $\sqrt{g^2+f^2-c}=\sqrt{a^2+b^2}$

Perpendicular distance = $\frac{|y_1-m{x}_1-c|}{\sqrt{1+m^2}}$

$y_1=b,x_1=a$ ....... ( Co-ordinates of center of circle)

$\therefore \sqrt{a^2+b^2}=\frac{|b-ma-c|}{\sqrt{1+m^2}}$

$\therefore \sqrt{a^2+b^2}\sqrt{1+m^2}=|b-ma-c|$

Opening mod,

$b-ma-c=\pm \sqrt{(a^2+b^2)(1+m^2)}$

$\therefore c=b-ma\pm \sqrt{(a^2+b^2)(1+m^2)}$