Question

Find the equation of the circle, the end points of whose diameter are (2, −3) and (−2, 4). Find its centre and radius.

Answer 1

(2, −3) and (−2, 4) are the ends points of the diameter of a circle. The equation of this circle is \[\left( x - 2 \right)\left( x + 2 \right) + \left( y + 3 \right)\left( y - 4 \right) = 0\]

\[\Rightarrow x^2 - 4 + y^2 - 4y + 3y - 12 = 0\]
\[ \Rightarrow x^2 + y^2 - y - 16 = 0 . . . (1)\]

Equation (1) can be rewritten as

\[x^2 + \left( y - \frac{1}{2} \right)^2 - \frac{1}{4} - 16 = 0\]
\[ \Rightarrow x^2 + \left( y - \frac{1}{2} \right)^2 = \frac{65}{4}\]

∴ Centre is \[\left( 0, \frac{1}{2} \right)\] and radius is\[\frac{\sqrt{65}}{2}\]