Coordinate Geometry: Question 8

Worked solution

Part (a): Equation of the circle

The radius is the distance from the centre $(3, -2)$ to the point $(7, 1)$ on the circle:

$r = \sqrt{(7 - 3)^2 + (1 - (-2))^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.$

Substituting centre $(a, b) = (3, -2)$ and $r = 5$ into $(x-a)^2 + (y-b)^2 = r^2$:

$(x - 3)^2 + (y - (-2))^2 = 5^2 \implies (x-3)^2 + (y+2)^2 = 25.$

So the equation of the circle is $\boxed{(x-3)^2 + (y+2)^2 = 25}$.

Part (b): Testing the point $D(-1, 2)$

Find the distance from the centre $(3, -2)$ to $D(-1, 2)$:

$CD = \sqrt{(-1 - 3)^2 + (2 - (-2))^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16 + 16} = \sqrt{32}.$

Compare this to the radius, $r = 5$ (note $r^2 = 25$). Since $32 > 25$, we have $\sqrt{32} > 5$, i.e. $CD > r$.

As a decimal check, $\sqrt{32} \approx 5.657$, which is indeed greater than $5$.

Since the distance from $D$ to the centre is greater than the radius, $D$ lies $\boxed{\text{outside}}$ the circle.

Part (c): Point diametrically opposite $(7, 1)$

If $(7, 1)$ and the unknown point $Q$ are the two ends of a diameter, the centre $(3, -2)$ must be the midpoint of $(7, 1)$ and $Q$. So:

$\left(\frac{7 + x_Q}{2}, \frac{1 + y_Q}{2}\right) = (3, -2).$

Solving for $Q$:

$\frac{7 + x_Q}{2} = 3 \implies 7 + x_Q = 6 \implies x_Q = -1$

$\frac{1 + y_Q}{2} = -2 \implies 1 + y_Q = -4 \implies y_Q = -5.$

Check: the distance from the centre $(3,-2)$ to $Q(-1,-5)$ is $\sqrt{(-1-3)^2+(-5-(-2))^2} = \sqrt{(-4)^2+(-3)^2} = \sqrt{16+9} = \sqrt{25} = 5$, equal to the radius, confirming $Q$ lies on the circle. $\checkmark$

So the point diametrically opposite $(7, 1)$ is $\boxed{(-1, -5)}$.

Final answers

• (a) $(x-3)^2 + (y+2)^2 = 25$
• (b) $D$ lies outside the circle ($CD = \sqrt{32} \approx 5.66 > 5$)
• (c) Diametrically opposite point: $(-1, -5)$