We are given two circles $S_1$($x^2+y^2-2x-3=0$) and $S_2$($x^2+y^2-4y-6=0$). A line $ax+by=2$ which touches the former circle and is normal to the latter. We have to find the value of a and b.
I was able to find the value of b:-
the center of $S_2$ is (-g,-f)=(0,2) Now since the line is normal to this circle, it passes through the circle :- $$y-2=m(x-0)$$ $$y=mx+2$$ So the value of b is 1. I don't know how to find the value of m.
On
Get the intersection points equation for the intersected circle. If you use the condition of getting single intersection point (one solution for quadratic equation) you should get two potential relations for a and b. Use the other relation, for the case of the line going through the center of the other circle, with the two relations you got from the intersection - two sets of two equations or one quadratic equation - solving this will provide you with two pairs of (a,b) solutions.
On
Since the line is normal to $S_2$ it must pass through the center of $S_2$, which is $C_2(0\mid 2)$ . Since the same line has to be tangent to $S_1$ , the problem boils down to constructing the tangent to $S_1$ from $C_2$ . The center of $S_1$ is $C_1(1\mid 0)$ . Construct the circle $S_3$ whose diameter is $C_2C_1$ . Its equation is $x^2+y^2-x-2y=0$ . Its intersections $P_1(-.6\mid 1.2),P_2(1\mid 2)$ with $S_1$ are the points of tangency. Draw $C_2P_1$ and $C_2P_2$ et voilà!
Plug in $ y = m x +2 $ into equation of circle S1:
$$ x^2 + ( m x+2)^2 - 2 x -3 =0 $$
$$ ( m^2 +1) x^2 + 2 x ( 2 m-1) + 1= 0$$
For tangency to S1 the roots must be same, or discriminant should vanish.
$$ \Delta =0, ( 2 m-1)^2- (m^2 +1) =0 $$
$$ m ( 3 m-4)= 0 $$
$$ m= (0, 4/3) $$
Thus there are 2 tangents that can be drawn whose slopes are $ (0,4/3) $ from center of S2.
Verify (0,2) as the intersection point of these 2 tangents.