Question

in the xy - plane, a circle has center c with coordinates (h,k). points a and b lie on the circle. point a has coordinates (h + 1,k + \sqrt{102}), and \angle acb is a right angle. what is the length of \overline{ab}?
(a) \sqrt{206}
(b) 2\sqrt{102}
(c) 103\sqrt{2}
(d) 103\sqrt{3}

Explanation:

Step1: Calculate the length of radius AC

Use the distance - formula $d=\sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}$. The coordinates of $C(h,k)$ and $A(h + 1,k+\sqrt{102})$. Then $AC=\sqrt{(h + 1 - h)^2+(k+\sqrt{102}-k)^2}=\sqrt{1^2+(\sqrt{102})^2}=\sqrt{1 + 102}=\sqrt{103}$.

Step2: Use the property of a right - angled isosceles triangle in a circle

Since $\angle ACB = 90^{\circ}$ and $AC = BC$ (radii of the same circle), by the Pythagorean theorem $AB^{2}=AC^{2}+BC^{2}$. And because $AC = BC=\sqrt{103}$, we have $AB^{2}=103+103 = 206$. Then $AB=\sqrt{206}$.

Answer:

A. $\sqrt{206}$