Question

What is the equation of this circle in standard form?

Answer 1

Answer:

Option C.

Step-by-step explanation:

Note : In the given points one point is (8,-2) instead of (-6,-2).

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$

where, (h,k) is center of circle and r is radius.

It is given that center of the circle is (-1,-2). So,

$h=-1,k=-2$

$(x-(-1))^2+(y-(-2))^2=r^2$

$(x+1)^2+(y+2)^2=r^2$    ...(1)

It is given that the circle passing through the point (8,-2),(-1,5),(6,-2),(-1,-9).

Substitute x=6 and y=-2 in equation (1).

$(6+1)^2+(-2+2)^2=r^2$

$(7)^2+(0)^2=r^2$

$49=r^2$

Substitute $r^2=49$ in equation (1).

$(x+1)^2+(y+2)^2=49$

Therefore, the correct option is C.