Question

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​

Answer 1

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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Answer 2

Step-by-step explanation:

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