Find equation of a circle

Question

Find equation of a circle passing through $(1, 1)$ and touching the circle $$ x^2 + y^2 + 4x - 6y - 3=0 $$ at the point $(2, 3)$. I am stuck as I cannot find more than $2$ equations for $3$ unknowns. I obtained my $2$ equations by substituting the two given points in general equation of circle. I don't understand how to make use of the equation of the other given circle.

Answer 1

So, the required circle passes through the intersection of the circle & its tangent at $(2,3)$

Find the equation of the tangent in the form $: Ax+By+C=0$

Now, the equation of any circle passing through the intersection of the given circle & $Ax+By+C=0$

can be written as $x^2+y^2+4x-6y-3+K(Ax+By+C)=0$ where $K$ is an arbitrary constant

Use the fact that this circle passes through $(1,1)$ to find $K$

Answer 2

Note that your equation is equivalent to $(x+2)^2+(y-3)^2=16$ hence it describes the circle with center $C(-2,3)$ and radius 4. Let $D(2,3)$ be the intersection point. Notice that $CD$ is parallel to the x-axis so the tangent at point $D$ must be parallel to the y-Axis and thus satisfies the equation $y=2$. Now, if the second circle touches this line also in D, then its center C' must be also with y-coordinate 3. Hence, the circle is symmetric to the line $y=3$ and you know that $(2,3),(1,1)$ and therefore also $(1,5)$ lies on the circle. Now you got three points for your circle and can set this into the general circle equation $(x-a)^2+(y-b)^2=r^2$ to obtain $r,a,b$.

Answer 3

So, first, you realize that the equation of the circle must be of the form $$ (x-h)^2 + (y-k)^2 = r^2. $$Then, since $(1,1)$ is on the circle, we get $$ (1-h)^2 + (1-k)^2 = r^2. $$Then, we note that $(2,3)$ is the only intersection of $$ (x+2)^2 + (y-3)^2 = 16 $$ and $$ (x-h)^2 + (y-k)^2 = r^2. $$So let us say we were given $h,k,r$. Then, we can solve for the intersection(s) of the two circles. We want $(2,3)$ to be the only one. One way to finish from here is to subtract the two equations, solve for $y$ in terms of $x$, substitute back in, and keep bashing.

However, there's an easier way. We want the two circles to be tangent at $(2,3)$, so that already brings it down to two circles $-$ one that is internally tangent and one that is externally tangent. In both cases, it is clear that we want the points $(-2,3)$, $(2,3)$, and $(h,k)$ collinear. This is just because the centers of the circle and the tangency point must be collinear.

Now, you can finish.