Question

the centre of a circle is (x+2,x-5), find the value of x , given that the circle passes through the points (2,-2)and(8,-2)


solve this​

Answer 1

Given ,

• Centre of a circle is O(x + 2 , x - 5)

• The circle passes through the points A(2 , -2) and B(8 , -2)

We know that , the distance between two points is given by

$\boxed{ \tt{Distance = \sqrt{ {( x_{2} - x_{1} )}^{2} + {(y_{2} - y_{1} )}^{2} } }}$

Thus ,

$\tt \implies OA = \sqrt{ {(2 - \{ x + 2)\})}^{2} + {( - 2 - \{x - 5 \} )}^{2} }$

$\tt \implies OA = \sqrt{ {(x)}^{2} + {( 3 - x)}^{2} }$

Similarly ,

$\tt \implies OB = \sqrt{ {(8 - \{ x + 2\})}^{2} + {( - 2 - \{ x - 5\})}^{2} }$

$\tt \implies OB = \sqrt{ {(x + 6)}^{2} + {( 3 - x)}^{2} }$

Since , OA and OB are radius

$\tt \implies\sqrt{ {(x)}^{2} + {( 3 - x)}^{2} } = \sqrt{ {(x + 6)}^{2} + {( 3 - x)}^{2} }$

Squaring on both sides , we get

$\tt \implies {(x)}^{2} + {( 3 - x)}^{2} = {(x + 6)}^{2} + {( 3 - x)}^{2}$

$\tt \implies {(x)}^{2} = {(x)}^{2} + {(6)}^{2} + 2(x)(6)$

$\tt \implies 0 = 36 + 12x$

$\tt \implies x = \frac{ - 36}{12}$

$\tt \implies x = - 3$

Therefore , the value of x is -3

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