Finding the equation of a circle given two points on the circle

Question

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.

Answer 1

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.

Answer 2

let the circle $$ (x-3)^2+ (y+1)^2 = 5$$ has a tangent at $(4, -3)$ with the circle $C.$ the line connecting the centers of the circle has slope $$\frac{-1-(-3)}{3-4} = -2$$.therefore the center of $C$ is at $(3+t, -1-2t)$ for some $t.$

since both $(4,-3), (0,7)$ are on $C,$ equating the radius squared, we have $$(3+t-4)^2 (-1-2t+3)^2 =(3+t-0)^2 +(-1-2t-7)^2 $$ that gives $$-8(3+t)+16 + -6(1+2t)+9=14(1+2t)+49\to 48t=-68, t = -17/7 $$

$$\text{ center of } C \text{ is }(4/7, 27/7), \text{ radius is } 7.666$$

Answer 3

I've come back to this question after a while and have found a solution which agrees with that in the textbook. My method is based primarily on the tips given by @Mufasa.

Let the circle which touches $x^{2} + y^{2} − 6x + 2y + 5 = 0$ at $(4,−3)$ and passes through $(0,7)$ be $C_{1}$.

Let the centre of $C_{1}$ be $(p, q)$ and let the radius be $a$.

$(4 - p)^{2} + (-3 - q)^{2} = a^{2}$

$p^{2} + (7 - q)^{2} = a^{2}$

Let the circle $x^{2} + y^{2} − 6x + 2y + 5 = 0$ be $C_{2}$.

The centre of $C_{2}$ is $(3, -1)$.

The gradient of the radius of $C_{2}$ to $(4, -3)$ is $-2$.

$\therefore$ The gradient of the tangent to $C_{2}$ at $(4, -3)$ is $\frac{1}{2}$.

$\therefore$ The gradient of the tangent to $C_{1}$ at $(4, -3)$ is $\frac{1}{2}$.

$\therefore$ The gradient of the radius of $C_{1}$ to $(4, -3)$ is $-2$.

$\therefore \dfrac{q - (-3)}{p - 4} = -2$

$2p + q - 5 = 0 \qquad (1)$

$(4 - p)^{2} + (-3 - q)^{2} = a^{2} \qquad (2)$

$p^{2} + (7 - q)^{2} = a^{2} \qquad (3)$

Equating (2) and (3), we ultimately get:

$p = \dfrac{5}{2}q - 3 \qquad (4)$

Substituting $p$ into (1), we get:

$q = \dfrac{11}{6}$

Substituting $q$ into (4), we get:

$p = \dfrac{19}{12}$

Substituting $p$ and $q$ into (3) we ultimately get:

$a^{2} = \dfrac{4,205}{144}$

This gives us our equation of $C_{1}$:

$$\left(x - \frac{19}{12}\right)^{2} + \left(y - \frac{11}{6}\right)^{2} = \frac{4,205}{144}$$

Expanding and simplifying, we get:

$$6x^{2} + 6y^{2} - 19x - 22y - 140 = 0.$$