Question

Show that $(3,3)$ is the centre of the circle passing through the points $(4,6)$, $(0,4)$, $(6,2)$ and $(4,0)$.

And what is the radius of the circle?

• A. $\sqrt{10}$ units
• B. $5$ units
• C. $\sqrt{12}$ units
• D. $4$ units

Answer 1

The correct option is A $\sqrt{10}$ units

We need to check if the distance between the centre and the points on the circle is the same.

Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ can be calculated using the formula $\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
Distance between the points O $(3,3)$ and A $(4,0)=\sqrt{{(4-3)}^2+{(0-3)}^2}=\sqrt{1+9}=\sqrt{10}$

Dince between the points O $(3,3)$ and B $(6,2)=\sqrt{{(6-3)}^2+{(2-3)}^2}=\sqrt{9+1}=\sqrt{10}$

Distance between the points O $(3,3)$ and C $(4,6)=\sqrt{{(4-3)}^2+{(6-3)}^2}=\sqrt{1+9}=\sqrt{10}$

Distance between the points O $(3,3)$ and D $(0,4)=\sqrt{{(0-3)}^2+{(4-3)}^2}=\sqrt{9+1}=\sqrt{10}$

Since, length of the sides between the centre and the all the other vertices are equal, they all lie on the circle with radius $\sqrt{10}$.