Question

Write the standard form of the equation of a circle with the given characteristics. Center: (– 8, – 3); Radius: 2√3

Answer 1

Given :

• Center: (– 8, – 3)

• Radius: 2√3

To find :

• equation of a circle

Formula used :

If center (x1 , x2) and radius (r) of circle is given then equation of circle :-

$\bf {(x - x_1)}^{2} + {(y - y_1)}^{2} = {r}^{2}$

Solution :

Center of circle = (– 8, – 3)

Radius = 2√3

$\sf \implies{(x - x_1)}^{2} + {(y - y_1)}^{2} = {r}^{2}$

$\sf \implies{(x + 8)}^{2} + {(y + 3)}^{2} = {(2 \sqrt{3}) }^{2}$

We know that :-

$\boxed{\sf {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab}$

$\sf \implies {x}^{2} + 64 + 16x + {y}^{2} + 9 + 6y = 12$

$\sf \implies {x}^{2} + 16x + {y}^{2} + 6y + 73 = 12$

$\sf \implies {x}^{2} + 16x + {y}^{2} + 6y + 73 - 12 = 0$

$\sf \implies {x}^{2} + {y}^{2} + 16x + 6y + 6 1= 0$

Answer :

$\sf equation \: of \: a \: circle : {x}^{2} + {y}^{2} + 16x + 6y + 6 1= 0$