In this page center and radius we are going to see example problems to find the equation of a circle with center and radius is given.
Example 1:
Find the equation of the circle if the center is (2,-3) and radius is 4 units
Solution:
The equation of the circle is (x-h)² + (y-k)² = r²
(h,k) = (2,-3) and r = 4
Equation of a circle:
(x-h)² + (y-k)² = r²
(x-2)² + (y-(-3))² = 4²
(x-2)² + (y+3)² = 16
x² + 2² - 2(x)(2) + y² + 3² + 2(y)(3) = 16
x² + 4 - 4x + y² + 9 + 6y = 16
x² - 4x + y² + 6y + 13 - 16 = 0
x² - 4x + y² + 6y -3 = 0
Example 2:
Find the equation of the circle if the center (1,5) and radius 9 and whether the circle passes through the point (2,0)
Solution:
The equation of the circle is (x-h)² + (y-k)² = r²
(h,k) = (1,5) and r = 9
Equation of a circle:
(x-h)² + (y-k)² = r²
(x-1)² + (y-5)² = 9²
x² + 1² - 2(x)(1) + y² + 5² + 2(y)(5) = 81
x² + 1 - 2x + y² + 25 + 10y = 81
x² - 2x + y² + 25y + 26 - 81 = 0
x² - 2x + y² + 25y -55 = 0
To check whether the circle passes through the point (2,0) we have to apply the values x=2 and y=0 in the equation of a circle.
2² - 2(2) + 0² + 25(0) -55 = 0
4 - 4 -55 = 0
-55 ≠ 0
So the circle is not passing through the point (2,0).
Example 3:
Find the equation of the circle if the center is (1,-3) and passing through the point (4,1)
Solution:
The equation of the circle is (x-h)² + (y-k)² = r²
But here we have only center of the circle.We don't have the radius.To find tht let us make diagram ith the given details.
To find the radius of a circle we have to find the distance from O to A.
Distance of OA (or) radius of a circle
OA = √(x₂ - x₁)² + (y₂ - y₁)²
Here x₁ = 1 x₂ = 4 y₁ = -3 y₂ = 1
OA = √(4 - 1)² + (1 - (-3))²
OA = √(3)² + (1 +3)²
OA = √9 + 4²
OA = √9 + 16
OA = √25
OA = √5 x 5
OA = 5 units
Now we have center and radius
Equation of a circle:
(x-h)² + (y-k)² = r²
(x-1)² + (y-(-3))² = 5²
(x-1)² + (y+3)² = 25
x² + 1² - 2(x)(1) + y² + 3² + 2(y)(3) = 25
x² + 1 - 2x + y² + 9 + 6y = 25
x² - 2x + y² + 6y + 10 - 25 = 0
x² - 2x + y² + 6y 15 = 0
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