# Find the centre and radius of the circle $x^2+y^2-8x+10y-12 = 0$

Toolbox:
• Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle. Then, by the definition, | CP | = r . By the distance formula, we have, $(x-h)^2+(y-k)^2=r^2$
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Given: the equation of the circle, we need to rewrite it in the form $(x-h)^2+(y-k)^2=r^2$, so that we can determine the centre as (h,k) and radius as $r$
Given $x^2+y^2-8x+10y-12 = 0$
$\quad \quad x^2 - 2\times 4x +16 +y^2 - 2\times 5y + 25 -25 -16 - 12 = 0$
$\quad \quad (x-4)^2 + (y+5)^2 = 53$
$\Rightarrow$ centre = $(4,-5)$ and radius = $\sqrt {53}$