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Find the equation of the circle passing through (2,-1) having the centre at (2,3). |
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Answer» The equation of circle whose radius is r and centre is (2, 3) is \((x -2)^2 + (y -3)^2 = r^2\) It is passing through points (2, - 1). \(\therefore (2 - 2)^2 + (- 1 - 3)^2 = r^2\) ⇒ \(r^2 = (-4)^2 = 16\) \(\therefore r = 4\) \(\therefore \) Equation of circle is \((x -2)^2 + (y -3)^2 = 4^2\) ⇒ \((x -2)^2 +(y - 3)^2 = 16\) or \(x^2 + y^2 - 4x - 6y + 4 + 9= 16\) ⇒ \(x^2 + y^2 - 4x -6y = 3\). |
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