find the equation of a circle the endpoints of whose one diameter are the centres of the circles x^2+y^2+6x-14y+5=0 and x^2+y^2-4x+10y+7=0
Answers
(x + 1/2)² + (y - 1)² = (13/2)² or x² + y² + x - 2y - 41 = 0 Equation of circle
Step-by-step explanation:
x²+ y² + 6x- 14y + 5=0
=> (x + 3)² - 9 + (y - 7)² - 49 + 5 =0
=>(x + 3)² + (y - 7)² = 53
Center of circle = (-3 , 7)
x²+y²- 4x+10y+7 = 0
=> (x - 2)² - 4 + (y + 5)² - 25 + 7 =0
=> (x - 2)² + (y + 5)² = 22
center of circle = ( 2 , -5)
Two End of Diameter
(-3 , 7) ( 2 , -5)
Length of Diameter = √(-3 - 2)² + (7 -(-5))² = √ 25 + 144 = 13
Radius = 13/2
Mid point = ( -1/2 , 1)
Equation of Circle
(x + 1/2)² + (y - 1)² = (13/2)²
=> x² + x + 1/4 + y² + 1 - 2y = 169/4
=> x² + y² + x - 2y - 41 = 0
Learn More
Equation of circle touching x axis at origin and the line 4x -3y +24=0 are
https://brainly.in/question/8752653
Find the equation of a circle with
https://brainly.in/question/7295471
diameter of a circle has endpoints P(-10,-2) and Q(4,6)
https://brainly.in/question/10212303