Question

The equation of a circle in general form is ​ x2+y2+22x+14y−55=0 ​ . What is the equation of the circle in standard form?
(x+11)2+(y+7)2=121

(x+11)2+(y+7)2=225

(x+22)2+(y+14)2=225

(x+22)2+(y+14)2=121

Answer 1

$x^2+y^2+22x+14y-55=0 \\\\x^2+22x+y^2+14y=55\\\\(x^2+22x+121)-121+(y^2+14y+49)-49=55\\\\(x+11)^2+(y+7)^2-121-49=55\\\\ (x+11)^2+(y+7)^2-170=55\\\\(x+11)^2+(y+7)^2=55+170\\\\\boxed{(x+11)^2+(y+7)^2=225}$

Answer B.

Answer 2

Answer:  $(x+11)^2+(y+7)^2=225$

Step-by-step explanation:

Since, the standard form of the circle is,

(x-h)² + (y-k)² = r²

Where (h,k) is the center of the circle and r be the radius of the circle.

Here, the given general form of the circle is,

$x^2+y^2+22x+14y-55=0$

$x^2+22x+y^2+14y=55$

$x^2+22x+11^2+y^2+14y+7^2=55+11^2+7^2$

$(x+11)^2+(y+7)^2=55+121+49$

$(x+11)^2+(y+7)^2=225$

Which is the required standard form of the circle.