Question

What is the equation of this circle in standard form?

(x−1)2+(y−2)2=7

(x+1)2+(y+2)2=7

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

(x−1)2+(y−2)2=49
A circle on a coordinate plane centered at begin ordered pair negative 1 comma negative 2 end ordered pair. The horizontal x-axis ranges from negative 10 to 10 increments of 1. The vertical y-axis ranges from negative 10 to 10 in increments of 1. The circle passes through begin ordered pair negative 6 comma negative 2 end ordered pair, begin ordered pair negative 1 comma 5 end ordered pair, begin ordered pair 6 comma negative 2 end ordered pair, and begin ordered pair negative 1 comma negative 9 end ordered pair.

Answer 1

Option C.

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.