Question

Find the equation of the circle with center at (−3,5) and passes through the point (5,−1) a. (x+3)² + (y-5)² = 100 b. (x-3)² + (y-5)² c. (x+3)² + (y-5)² d. None of the above

Answer 1

a

Explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

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

given centre of circle is (- 3, 5 ) , then

(x - (- 3) )² + (y - 5)² = r² , that is

(x + 3)² + (y - 5)² = r² → **

the radius r is the distance from the centre to any point on the circle.

given the circle passes through the point (5, - 1)

We can find r using the distance formula

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

let (x₁, y₁ ) = (- 3, 5 ) and (x₂, y₂ ) = (5, - 1 )

r =
`√((5-(-3))^2+(-1-5)^2)`

=
`√((5+3)^2+(-6)^2)`

=
`√(8^2+36)`

=
`√(64+36)`

=
`√(100)`

r = 10 ( square both sides )

r² = 10² = 100

return to **

(x + 3)² + (y - 5)² = 100 ← equation of circle