find the equation of the circle the end points of whose diameter are (-3,7) and (5,1) also find the length of intercept made by it on x axis
Answers
equation of circle , x² + y² - 2x - 8y -8 = 0
x-intercepts ; 6 unit
we have to find the equation of circle end points of whose diameter are A(-3, 7) and B(5,1).
first find centre of circle by using midpoint section formula,
centre , O = [(-3 + 5)/2, (7 + 1)/2]
= (1, 4)
now, radius of the circle , r = OA = OB
r = OA = √{(1 + 3)² + (4 - 7)²}
= √{(4)² + (-3)²}
= √(16 + 9)
= √25
= 5
equation of circle is given as
(x - a)² + (y - b)² = r²
where (a, b) is co-ordinate of the centre of circle.
so, (x - 1)² + (y - 4)² = 5²
⇒x² - 2x + 1 + y² - 8y + 16 = 25
⇒x² + y² - 2x - 8y - 8 = 0 [ this is required equation of circle ]
now length of intercept made by it on x - axis ;
let y = 0 then, equation x² + 0² - 2x - 8(0) - 8 = 0
x² - 2x - 8 = 0
x² - 4x + 2x - 8 = 0
x = -2, 4
hence , (-2,0) and (4,0) are x-intercepts
so, length of x-intercepts = √{(4 + 2)² + (0)²} = 6
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