Question

The centre of a circle is (x+2, x-1). Find r if the circle passes through (2-2) and
(8-2).​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

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Answer:

Given that, a circle passes through points ( 2, -2 ) and ( 8, -2 ).

This implies that the circle's outer circumference passes through these points.

We know that, the distance from circle to any point on the circumference is equal to each other as it is called as radius.

So distance between both the points from the center is equal.

So calculating distance from center for point ( 2, -2 ) we get,

=> √ ( x + 2 - 2 )² + ( x - 1 - ( -2 ) )²  = Radius

=> √ ( x + 0 )² + ( x - 1 + 2 )² = Radius

=> √ x² + ( x + 1 )² = Radius   ... ( 1 )

Now calculating the distance between Center to ( 8, -2 ) we get,

=> √ ( x + 2 - 8 )² + ( x - 1 - ( -2 ) )² = Radius

=> √ ( x - 6 )² + ( x - 1 + 2 )² = Radius

=> √ ( x - 6 )² + ( x + 1 )² = Radius   ... ( 2 )

Equating the LHS of ( 1 ) and ( 2 ) we get,

=>  √ x² + ( x + 1 )² = √ ( x - 6 )² + ( x + 1 )²

Squaring on both sides, we cancel the roots. Hence we get,

=> x² + ( x + 1 )² = ( x - 6 )² + ( x + 1 )²

( x + 1 )² gets cancelled and we get,

=> x² = ( x - 6 )²

=> x² =  x² - 12x + 36

=> x² - x² + 12x = 36

=> 12x = 36

=> x = 36 / 12 = 3

Hence the value of x is 3.

Centre coordinates: ( 3 + 2, 3 - 1 )

=> Centre Coordinates: ( 5, 2 )

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