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Find the coordinates of the centre of the circle passing through the three given points A(5, 1), B(-3, -7) and C(7, -1).

Coordinate Geometry

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Answer

Let O(x, y) be the coordinates of the centre of the circle. Points A(5, 1), B(-3, -7), and C(7, -1) are on the circle.

Find the coordinates of the centre of the circle passing through the three given points A(5, 1), B(-3, -7) and C(7, -1). Coordinate Geometry, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

By distance formula,

d = (x2−x1)2+(y2−y1)2\sqrt{(x2 - x1)^2 + (y2 - y1)^2}

As, OA = OB [∵ Both are radius of the same circle]

∴(x−5)2+(y−1)2=[x−(−3)]2+[y−(−7)]2⇒x2+25−10x+y2+1−2y=[x+3]2+[y+7]2⇒x2+y2−10x−2y+26=x2+9+6x+y2+49+14y⇒x2+y2−10x−2y+26=x2+y2+6x+14y+58\therefore \sqrt{(x - 5)^2 + (y - 1)^2} = \sqrt{[x - (-3)]^2 + [y - (-7)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 25 - 10x + y^2 + 1 - 2y} = \sqrt{[x + 3]^2 + [y + 7]^2} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 10x - 2y + 26} = \sqrt{x^2 + 9 + 6x + y^2 + 49 + 14y} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 10x - 2y + 26} = \sqrt{x^2 + y^2 + 6x + 14y + 58}

On squaring both sides,

⇒x2+y2−10x−2y+26=x2+y2+6x+14y+58⇒x2−x2+y2−y2+6x+10x+14y+2y=26−58⇒16x+16y=−32⇒16(x+y)=−32⇒x+y=−3216⇒x+y=−2⇒x=−2−y .........(1).\Rightarrow x^2 + y^2 - 10x - 2y + 26 = x^2 + y^2 + 6x + 14y + 58 \\[1em] \Rightarrow x^2 - x^2 + y^2 - y^2 + 6x + 10x + 14y + 2y = 26 - 58 \\[1em] \Rightarrow 16x + 16y = -32 \\[1em] \Rightarrow 16(x + y) = -32 \\[1em] \Rightarrow x + y = \dfrac{-32}{16} \\[1em] \Rightarrow x + y = -2 \\[1em] \Rightarrow x = -2 - y \space ………(1).

As, OC = OB [∵ Both are radius of the same circle]

∴(x−7)2+[y−(−1)]2=[x−(−3)]2+[y−(−7)]2⇒x2+49−14x+[y+1]2=[x+3]2+[y+7]2⇒x2+49−14x+y2+1+2y=x2+9+6x+y2+49+14y⇒x2+y2−14x+2y+50=x2+y2+6x+14y+58\therefore \sqrt{(x - 7)^2 + [y - (-1)]^2} = \sqrt{[x - (-3)]^2 + [y - (-7)]^2} \\[1em] \Rightarrow \sqrt{x^2 + 49 - 14x + [y + 1]^2} = \sqrt{[x + 3]^2 + [y + 7]^2} \\[1em] \Rightarrow \sqrt{x^2 + 49 - 14x + y^2 + 1 + 2y} = \sqrt{x^2 + 9 + 6x + y^2 + 49 + 14y} \\[1em] \Rightarrow \sqrt{x^2 + y^2 - 14x + 2y + 50} = \sqrt{x^2 + y^2 + 6x + 14y + 58}

On squaring both sides,

⇒x2+y2−14x+2y+50=x2+y2+6x+14y+58⇒x2+y2−x2−y2+6x+14x+14y−2y=50−58⇒20x+12y=−8⇒5x+3y=−2\Rightarrow x^2 + y^2 - 14x + 2y + 50 = x^2 + y^2 + 6x + 14y + 58 \\[1em] \Rightarrow x^2 + y^2 - x^2 - y^2 + 6x + 14x + 14y - 2y = 50 - 58 \\[1em] \Rightarrow 20x + 12y = -8 \\[1em] \Rightarrow 5x + 3y = -2 \\[1em]

Substituting value of x in above equation from equation 1 :

⇒ 5(-2 - y) + 3y = -2

⇒ -10 - 5y + 3y = -2

⇒ -2y = -2 + 10

⇒ -2y = 8

⇒ y = -4.

⇒ x = -2 - y = -2 - (-4) = -2 + 4 = 2.

C = (x, y) = (2, -4).

Hence, the coordinates of the centre of the circle are (2, -4).

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