Coordinate Geometry: Question 9

Worked solution

Part (a): Equation of the circle

Centre. Since $GH$ is a diameter, the centre is the midpoint of $G(-3, 4)$ and $H(5, -2)$:

$\left(\frac{-3 + 5}{2}, \frac{4 + (-2)}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1).$

Radius. First find the full diameter length $GH$:

$GH = \sqrt{(5 - (-3))^2 + (-2 - 4)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10.$

The radius is half of this: $r = 10 \div 2 = 5$.

Equation. Substituting centre $(1, 1)$ and $r = 5$ into $(x-a)^2+(y-b)^2=r^2$:

$(x - 1)^2 + (y - 1)^2 = 25.$

Check: the distance from the centre $(1,1)$ to $G(-3,4)$ is $\sqrt{(-3-1)^2+(4-1)^2}=\sqrt{16+9}=\sqrt{25}=5$, equal to $r$. $\checkmark$

So the circle’s equation is $\boxed{(x-1)^2 + (y-1)^2 = 25}$.

Part (b): $GK \perp KH$ using the semicircle property

Since $GH$ is a diameter, the circle theorem “the angle in a semicircle is $90°$” tells us that for any point $K$ on the circle (other than $G$ or $H$), the angle $\angle GKH = 90°$. Equivalently, $GK$ must be perpendicular to $KH$. We verify this using gradients.

Gradient of $GK$, between $G(-3, 4)$ and $K(4, 5)$:

$m_{GK} = \frac{5 - 4}{4 - (-3)} = \frac{1}{7}.$

Gradient of $KH$, between $K(4, 5)$ and $H(5, -2)$:

$m_{KH} = \frac{-2 - 5}{5 - 4} = \frac{-7}{1} = -7.$

Product of gradients:

$m_{GK} \times m_{KH} = \frac{1}{7} \times (-7) = -1.$

Since the product of the gradients is $-1$, $GK$ is perpendicular to $KH$. $\blacksquare$

(As an independent check: the distance from the centre $(1,1)$ to $K(4,5)$ is $\sqrt{(4-1)^2+(5-1)^2}=\sqrt{9+16}=\sqrt{25}=5=r$, confirming $K$ does lie on the circle, consistent with the perpendicularity result.)

Part (c): Testing the point $L(6, 3)$

Find the distance from the centre $(1, 1)$ to $L(6, 3)$:

$CL = \sqrt{(6 - 1)^2 + (3 - 1)^2} = \sqrt{5^2 + 2^2} = \sqrt{25 + 4} = \sqrt{29}.$

Compare $CL^2 = 29$ with $r^2 = 25$. Since $29 \neq 25$, $CL \neq r$ (in fact $CL > r$, since $29 > 25$).

So $L$ does not lie on the circle. It lies outside it, since $\sqrt{29} \approx 5.39 > 5$.

Final answers

• (a) $(x-1)^2 + (y-1)^2 = 25$
• (b) $m_{GK} \times m_{KH} = \dfrac{1}{7} \times (-7) = -1$, so $GK \perp KH$
• (c) $L(6, 3)$ does not lie on the circle ($CL = \sqrt{29} \neq 5$)