Question

Find the centre of the circle passing through (-4,-11), (10,-13) and (-2, -13).
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Answer 1

Given ,

the circle passing through (-4,-11), (10,-13) and (-2, -13)

Let , the centre of circle be (x , y)

According to the question ,

$\tt \implies AO = \sqrt{ { \{x - ( - 4) \}}^{2} + { \{ y - ( - 11)\}}^{2} }$

$\tt \implies AO = \sqrt{ { \{ x + 4\}}^{2} + { \{ y + 11\}}^{2} }$

Similarly ,

$\tt \implies BO = \sqrt{ { \{x - 10\}}^{2} + { \{ y - ( - 13)\}}^{2} }$

$\tt \implies BO = \sqrt{ { \{ x - 10\}}^{2} + { \{ y + 13\}}^{2} }$

And

$\tt \implies CO = \sqrt{ { \{x - ( - 4) \}}^{2} + { \{ y - ( - 11)\}}^{2} }$

$\tt \implies CO = \sqrt{ { \{ x + 2\}}^{2} + { \{ y + 13\}}^{2} }$

Since , AO , BO and CO are the radius of circle

Thus ,

$\tt \implies AO = BO$

$\tt \implies \sqrt{ { \{ x + 4\}}^{2} + { \{ y + 11\}}^{2} } = \sqrt{ { \{ x - 10\}}^{2} + { \{ y + 13\}}^{2} }$

Squaring on both sides , we get

$\tt \implies { {(x + 4)}^{2} + (y + 11)}^{2} = {(x - 10)}^{2} + {(y + 13)}^{2}$

$\tt \implies {(x)}^{2} + 16 + 8x + {(y)}^{2} + 121 + 22y = {(x)}^{2} + 100 - 20x + {(y)}^{2} + 169 + 26y$

$\tt \implies 137 + 8x + 22y = 269 - 20x + 26y$

$\tt \implies 28x - 4y = 132$

$\tt \implies 7x - y = 33 - - - - (i)$

And

$\tt \implies BO = CO$

$\tt \implies \sqrt{ { \{ x - 10\}}^{2} + { \{ y + 13\}}^{2} } = \sqrt{ { \{ x + 2\}}^{2} + { \{ y + 13\}}^{2} }$

Squaring on both sides , we get

$\tt \implies {(x - 10)}^{2} + {(y + 13)}^{2} = {(x + 2)}^{2} + {(y + 13)}^{2}$

$\tt \implies {(x)}^{2} + 100 - 20x = {(x)}^{2} + 4 + 4x$

$\tt \implies 96 = 24x$

$\tt \implies x = \frac{96}{24}$

$\tt \implies x = 4$

Put the value of x = 4 in eq (i) , we have

$\tt \implies 7(4) - y = 33$

$\tt \implies y = 28 - 33$

$\tt \implies y = - 5$

Hence , the center of circle is (4 , -5)