Question

A circle is centered at the point (5, -4) and passes through the point (-3, 2).

The equation of this circle is (x + ?)2 + (y + ?)2 = ?

Answer 1

Hello,

r²=(5+3)²+(-4-2)²=8²+6²=100

(x-5)²+(y+4)²=100

Answer 2

`(x-5)^2+(y+4)^2=100`

Explanation:

The standard form of a circle is

`(x-h)^2+(y-k)^2=r^2` .... (1)

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

It is given that a circle is centered at the point (5, -4) and passes through the point (-3, 2).

`h=5,k=-4`

Distance between (5, -4) and (-3, 2) is radius of the circle.

`r=√((x_2-x_1)^2+(y_2-y_1)^2)`

`r=√((-3-5)^2+(2-(-4))^2)`

`r=√((-8)^2+(6)^2)`

On further simplification we get

`r=√(64+36)`

`r=√(100)`

`r=10`

The radius of the circle is 10 units.

Substitute h=5,k=-4 and r=10 in equation (1).

`(x-5)^2+(y-(-4))^2=(10)^2`

`(x-5)^2+(y+4)^2=100`

Therefore, the equation of the circle is
`(x-5)^2+(y+4)^2=100`.