I have to find the equation of the circle which passes through the points $A(-2,2)$ and $B(5,-5)$ and has the line $3x-4y=35$ as a tangent at the point $B(5,-5)$.
I tried forming the tangent in a third equation using $(-g,-f)$. With the two points on the circle I used them to form an equation in the form $x^2+y^2+2gx+2fy+c$. I don't know how to go further.
Best Answer
The equation of the line perpendicular to the given tangent at $B (5,-5)$ can be found to be $4x + 3y = 5$. The centre of the circle must lie on this line. The centre must also lie on the perpendicular bisector of the line joining $A$ and $B$. The equation of this perpendicular can be found by using the fact that it passes through the mid point of $AB$ and perpendicular to $AB$. The equation of this line will be $x - y = 3$. Solving the linear equation gives the centre as $(2,-1)$. The radius is $\sqrt{4^2+(-3)^2} = 5$.
So the equation of circle is $(x-2)^2 +(y+1)^2 = 25$ which can be expressed in standard form as $x^2 + y^2 - 4x + 2y -20 = 0$. This is the answer.