find the equation of a circle from 3 points on circumference

Question

The following question is from higher maths 2014 Scotland

(a) Find P and Q, the points of intersection of the line ${y = 3x - 5}$ and the circle ${C_1}$ with the equation ${x^2 + y^2 + 2x - 4y -15 =0}$.

(b) T is at the centre ${C_1}$. Show PT and QT are perpendicular.

(c) A second circle ${C_2}$ passes through P, Q and T. Find the equation of ${C_2}$.

I have solved a and b but I am stuck on part (c).

After parts a and b, I have 3 points on the circumference P (1, -2), Q (3, 4) and T (-1, 2).

I know that a perpendicular bisector to a chord will always pass through the centre.

So am I right in saying that I should find the perpendicular bisectors of PT and QT (I have the gradients from question b) and then solve the 2 equations simultaneously.

The perpendicular bisector of PT is

Midpoint PT = ${({1 + 1 \over 2}, {-2 + 2 \over 2})}$ = (0, 0)

gradient of PT = ${2 - -2 \over -1 - 1}$ = -2, perpendicular gradient = ${1 \over 2}$

equation of PT = ${y = 1\over2 x}$

=> ${2y -x = 0}$

Midpoint of QT = ${({3 + -1\over 2}, {2 + 4 \over 2})}$ = (1, 3)

gradient of QT = ${-2}$

equation of QT = ${y - 2 = -2(x + 1)}$

=> ${y - 2 = -2x -2}$

=> ${y + 2x = 0}$

I would then solve them simultaneously to find the point (0, 0) which seems wrong.

I take it I have taken a wrong term somewhere?

Answer 1

You have a mistake in finding the perpendicular bisector of $QT$.

The equation of the line, whose slope is $-2$, passing through $(1,3)$ is $$y-3=-2(x-1),$$ not $$y-2=-2(x+1).$$

Answer 2

An easy solution is achieved by translating the point $P$ to the origin (and $Q,T$ accordingly).

Then the equation of a circle by the origin is

$$ax+by=x^2+y^2.$$

Plugging the coordinates of $Q-P$ and $T-P$, you easily solve

$$ax_q+by_q=x_q^2+y_q^2\\ ax_t+by_t=x_t^2+y_t^2,$$

and the coordinates of the center are $\frac12(a,b)+P$, and the radius $\frac12\sqrt{a^2+b^2}$.

Answer 3

mathlove has pointed to your error, but here is a simpler way to proceed:

You proved in part (b) that $PT$ and $QT$ are perpendicular, so $PTQ$ is a right triangle, and the center of its circumcircle is the midpoint of the hypotenuse $PQ$.

(It would also be simple to see this by plotting the points on graph paper!)

Answer 4

First, put $C_1$ in standard form:

$$(x+1)^2+(y-2)^2=20$$

a) is algebra, easy to verify that $(3,4)$ and $(1, -2)$ are the two solutions.

b) is easiest, I think, by showing that the triangle $\Delta TPQ$ is $45-45-90$, which requires showing that $\overline{PQ}$ is $\sqrt{2}$ as long as the radius of $C_1$, i.e. $\sqrt{40}$, which it is (easy application of distance formula).

I think c) is easy because we know the circumcircle of a right triangle has its center on the midpoint of the hypotenuse (easy to see that's $(2, 1)$), and thus radius half the hypotenuse ($\frac{\sqrt{40}}2$). Which, in standard form, is:

$$(x-2)^2+(y-1)^2=10$$

(and blessedly these are all integers, so it's easy to verify that $T$, $P$, and $Q$ all satisfy this)

Answer 5

In general, the equation of the circle through the three points $(x_1, y_1)$, $(x_2, y_2)$, $(x_3, y_3)$ is $$ \left| \; \begin{matrix} x^2 + y^2 & x & y & 1 \\ x_1^2 + y_1^2 & x_1 & y_1 & 1 \\ x_2^2 + y_2^2 & x_2 & y_2 & 1 \\ x_3^2 + y_3^2 & x_3 & y_3 & 1 \\ \end{matrix} \;\right| = 0 $$