Question: A circle touches the lines $2x+3y+1=0$ at the point (1, -1) and is orthogonal to the circle which has the line segment having end points (0, -1) and (-2, 3) as the diameter. What is the equation of the circle?
Approach so far: Using the given coordinates of the line segment, the radius as well as the center of the circle can easily be calculated. However I am lost as how to proceed further. Please help!
There are 2 circles called $C_k$ and $C_r$.
$C_k$ is the known circle with center at … (-1, 1) and $radius^2 = r^2 = … = 20$.
$C_r$ is the required circle with center at (h, k) and radius = R = d[(h, k), (1, -1)].
The line perpendicular to $2x + 3y +1 = 0$ and passing through (1, -1) is …. $2x – 3y – 5 = 0$
(h, k) lies on $2x – 3y – 5 = 0$ implies $2h – 3k – 5 = 0$ …… (1)
By orthogonality, r^2 + R^2 = (distance between centers)^2
i.e. $20 + [d[(h, k), (1, -1)]]^2 = [d[(h, k), (-1, 1)]]^2$ ….. (2)
(2) can be reduced to a linear equation in h and k (I hope it is independent from (1).)
Solving (1) and (2) will give you the values of h and k. This in turn means C_r can be found.