Question

Determine the equation of the circle with center 100pts

Answer 1

`(x-8)^2+(y-5)^2=400`

Explanation:

The standard equation of a circle is:

`\boxed{(x-h)^2+(y-k)^2=r^2}`

where:

• (h, k) is the center.
• r is the radius.

The given center of the circle is (8, 5).

To find the value of r², substitute the circle and the given point (-4, 21) into the equation and solve for r².

`\begin{aligned}(-4-8)^2+(21-5)^2&=r^2\\(-12)^2+(16)^2&=r^2\\144+256&=r^2\\400&=r^2\end{aligned}`

Finally, substitute the center and r² into the formula to create an equation of the circle with the given parameters:

`\boxed{(x-8)^2+(y-5)^2=400}`

Answer 2

(x - 8)² + (y - 5)² = 400

Explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r the radius

the radius is the distance from the centre to a point on the circle

use the distance formula to calculate r

r =
`\sqrt{(x_(2)-x_(1))^2+(y_(2)-y_(1))^2 }`

with (x₁, y₁ ) = (8, 5 ) and (x₂, y₂ ) = (- 4, 21 )

r =
`√((-4-8)^2+(21-5)^2)`

=
`√((-12)^2+16^2)`

=
`√(144+256)`

=
`√(400)`

= 20

then with (h, k ) = (8, 5 ) and r = 20, the equation of the circle is

(x - 8)² + (y - 5)² = 20² , that is

(x - 8)² + (y - 5)² = 400