Show that $(4, 3)$ is the centre of the circle which passes through the points $(9, 3), (7, - 1) a n d (1, - 1)$ . Find also its radius.

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Solution

Suppose C represents the pints (4, 3). Let P, Q and R denote the points (9, 3), (7, -1) and (1, - 1) respectively. Using the distance formula
$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
we get
$C P^{2} = (9 - 4)^{2} + (3 - 3)^{2} = 5^{2} = 25$
$C Q^{2} = (7 - 4)^{2} + (- 1 - 3)^{2} = 3^{2} + (- 4)^{2} = 9 + 16 = 25$
$C r^{2} = (4 - 1)^{2} + (3 + 1)^{2} = 3^{2} + 4^{2} = 9 + 16 = 25$
So, $C P^{2} = C Q^{2} = C R^{2} = 25$ or $C P = C Q = C R = 5$ .
Hence, the points P, Q, R are on the circle with centre at (4, 3) and its radius is 5 units.

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