Find the equations of the circles that have centre $(0,0)$ and touch the circle $x^2 + y^2 - 8x - 6y + 24 = 0$
So far I have said:
As the circles have centre $(0,0)$ their equations are of the form $x^2 + y^2 = r^2$ where $r$ is the radius. Subbing this into the equation of the given circle gives $r^2 = 8x + 6y - 24$.
I have also said that the tangents to the circles at the points of contact have the same gradient. So differentiating the given circle's equation gives $\frac{dy}{dx} = \frac{8-2x}{2y-6}$ and differentiating $r^2 = x^2 + y^2$ gives $\frac{dy}{dx} = \frac{-x}{y}$. Then, setting these equal to each other gives $8y = 6x$.
From this point I'm not sure what to do. Any pointers? Thanks.
This does not need calculus, just a theorem from euclidean geometry. The given circle has centre $(4,3)$ and radius $1$ unit. Further distance of $(4,3)$ from $(0,0)$(centre of the other circle) is $5$ units. Let $r$ be its radius. Then we would have either $r+1=5$ (for touching externally) or $r-1=5$ (for touching internally). Thus there are two circles centred at the origin having radii $4$ or $6$ units. Now you can find the equations.