I have to find the locus of centres of circles that pass through the point $(8,0)$ and tangent to circle $x^2+y^2=100$
$(x_0, y_0)$ – centres we are looking for
I decided to try something with distances, so I made a the equation $(x_0^2-8)^2+y_0^2=r^2$ It is the distance from center to point
Also I found the equation for tangent for our big circle $x_1x+y_1y-100=0$, where $x_1, y_1$ a point on the circle
Equatation for distance from center to tangent is $\frac {(x_1x_0+y_1y_0-100)^2}
{x_1^2+y_1^2} =r^2$
But I dont have any results
Best Answer
Note that $P = (8,0)$ is located inside the circle $O: x^2 + y^2 = 10^2$, which is centered at the origin and has radius $10$. A circle tangent to $O$ and passing through $P$ consequently will be internally tangent to $O$. Suppose such a circle, which we will label $Q$, has center $(x_Q, y_Q)$; then $(x_Q, y_Q)$ is equidistant from $P$ and the point of tangency of $Q$ to $O$, and this distance is the radius of $Q$, which we will denote $r_Q$. That is to say, $$r_Q^2 = (x_Q - 8)^2 + (y_Q - 0)^2 = \left(10 - \sqrt{x_Q^2 + y_Q^2}\right)^2.$$ The rightmost expression arises from the fact that the point of tangency of two circles lies on the line joining their centers; thus the distance of $(x_Q, y_Q)$ to the point of tangency of $Q$ to $O$ is equal to the radius of $O$ minus the distance of $(x_Q, y_Q)$ to the origin.
Simplifying the RHS equality then gives $$(x_Q - 8)^2 + y_Q^2 = 10^2 - 20 \sqrt{x_Q^2 + y_Q^2} + x_Q^2 + y_Q^2,$$ or $$16x_Q + 36 = 20 \sqrt{x_Q^2 + y_Q^2},$$ or $$(4x_Q + 9)^2 = 25(x_Q^2 + y_Q^2),$$ or $$9x_Q^2 - 72x_Q + 25y_Q^2 = 81,$$ Completing the square gives $$9(x_Q - 4)^2 + 25y_Q^2 = 225,$$ or in standard form, $$\frac{(x_Q - 4)^2}{5^2} + \frac{y_Q^2}{3^2} = 1.$$ This is an ellipse with center $(4,0)$, major semiaxis $5$, and minor semiaxis $3$.