Find the centre and radius of the circle given by the equation: $x^{2}$ + $y^{2}$ + 26x + 34y + 97 = 0.

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Solution

Rearranging the equation,
$x^{2}$ + $y^{2}$ + 26x + 34y + 97 = 0
$x^{2}$ + 26x + $y^{2}$ + 34y + 97 = 0
( $x^{2}$ + 2 $\times$ 13 $\times$ x + $13^{2}$ ) - 169 + ( $y^{2}$ + 2 $\times$ 17 $\times$ y + $17^{2}$ ) - 289 + 97 = 0
$(x + 13)^{2}$ + $(y + 17)^{2}$ - 361 = 0
$[x - (- 13)]^{2}$ + $[y - (- 17)]^{2}$ = $19^{2}$

So the circle's centre is at (-13, -17) and its radius is 19 units.

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Find the centre and radius of the circle given by the equation: x2 + y2 + 26x + 34y + 97 = 0.

Rearranging the equation, x2 + y2 + 26x + 34y + 97 = 0 x2 + 26x + y2 + 34y + 97 = 0 (x2 + 2×13×x + 132) - 169 + (y2 + 2×17×y + 172) - 289 + 97 = 0 (x+13)2 + (y+17)2 - 361 = 0 [x−(−13)]2 + [y−(−17)]2 = 192 So the circle's centre is at (-13, -17) and its radius is 19 units.

Find the centre and radius of the circle given by the equation: $x^{2}$ + $y^{2}$ + 26x + 34y + 97 = 0.