11. Find the equation of the circle which touches $x^{2} + y^{2} – 6x + 2y + 5 = 0$ at $(4, -3)$ and passes through $(0, 7)$.
My textbook has a worked example for obtaining the equation of a circle from three points on the circle. It also talks you through obtaining the equation of a circle if you're given two points on the circle and if it touches an axis (in this case you know a coordinate of the centre will be $\pm r$, where $r$ is your radius.)
I am reasonably well-practised at using these two techniques. I have also been practising finding the length of a tangent from a given point.
However, I have so far been unable to make the "jump" to this question, I suspect there's something I'm not seeing. So I was hoping for a hint that would help me work out how to approach this question.
Best Answer
Lets say the equation of the circle is given by:$$(x-a)^2+(y-b)^2=r^2$$Make use of the fact that the circle passes through $(4,-3)$ and $(0,7)$ to form two equations.
You are also told that it touches the circle:$$x^{2} + y^{2} - 6x + 2y + 5 = 0$$at $(4,-3)$ which mean the tangents of both circles at this point must be equal. This gives you a third equation.
You now have three equations and three unknowns which you should be able to solve.