Question

𝐴(1,−2) is a point on the circle (𝑥−3)2+ (𝑦+1)2= 5

a. State the coordinates of the centre of the circle and hence find the coordinates of the point 𝐵 where 𝐴𝐵 is the diameter of the circle.

b. 𝐶(2,1) also lies on the circle. Use coordinate geometry to verify that angle 𝐴𝐶𝐵 = 900

Answer 1

a.

$$(𝑥−3)^2+ (𝑦+1)^2= 5$$

The coordinates of the centre of the circle are $D(3, -1).$

If $AB$ is the diameter of the circle, then $D$ is the midpoint of $AB$

$$x_D=\dfrac{x_A+x_B}{2}=>x_B=2x_D-x_A$$

$$y_D=\dfrac{y_A+y_B}{2}=>y_B=2y_D-y_A$$

$$x_B=2(3)-1=5, y_B=2(-1)-(-2)=0$$

Point $B(5, 0)$

b.

Point $C(2, 1)$

$$\overrightarrow{CA}=\langle1-2, -2-1\rangle$$

$$\overrightarrow{CB}=\langle5-2, 0-1\rangle$$

$$\overrightarrow{CA}\cdot \overrightarrow{CB}=-1(3)+(-3)(-1)=-3+3=0$$

$$|\overrightarrow{CA}|\not=0, |\overrightarrow{CB}|\not=0$$

Therefore $\overrightarrow{CA}\perp \overrightarrow{CB},$ and $\angle ACB=90\degree.$