Find the radius and centre of the circle given by the equation $x^{2}$ + $y^{2}$ + 22x + 40y + = 0.

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Solution

Rearranging the equation,
$x^{2}$ + $y^{2}$ + 22x + 40y + 37 = 0
$x^{2}$ + 22x + $y^{2}$ + 40y + 37= 0
( $x^{2}$ + 2 $\times$ 11 $\times$ x + $11^{2}$ ) - 121 + ( $y^{2}$ + 2 $\times$ 20 $\times$ y + $20^{2}$ ) - 400 +37 = 0
$(x + 11)^{2}$ + $(y + 20)^{2}$ - 484 = 0
$[x - (- 11)]^{2}$ + $[y - (- 20)]^{2}$ = $22^{2}$

So the circle's centre is at (-11, -20) and its radius is 22 units.

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Find the radius and centre of the circle given by the equation x2 + y2 + 22x + 40y + = 0.

Rearranging the equation, x2 + y2 + 22x + 40y + 37 = 0 x2 + 22x + y2 + 40y + 37= 0 (x2 + 2×11×x + 112) - 121 + (y2 + 2×20×y + 202) - 400 +37 = 0 (x+11)2 + (y+20)2 - 484 = 0 [x−(−11)]2 + [y−(−20)]2 = 222 So the circle's centre is at (-11, -20) and its radius is 22 units.

Find the radius and centre of the circle given by the equation $x^{2}$ + $y^{2}$ + 22x + 40y + = 0.