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Find the equation of the circle with radius 5 whose center lies on x-axis and passes through the point (2, 3). |
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Answer» Let the equation of the circle be (x – h)2 + (y – k)2 = r2 center lies on x axis so let the center be (h, 0), then (x – h)2 + y2 = 25 Since circle pass through (2, 3) we have; (2 – h)2 + 32 = 25 ⇒ (2 – h)2 = 16 ⇒ 2 – h = ±4 ⇒ h = 6, -2 When h = 6 ; equation of circle is (x – 6)2 + (y – 0)2 = 25 ⇒ x2 + 36 – 12x + y2 = 25 ⇒ x2 + y2 – 12x + 11 = 0 When h = -2; equation of circle is (x + 2)2 + (y – 0)2 = 25 ⇒ x2 + 4 + 4x + y2 = 25 ⇒ x2 + y2 + 4x – 21 = 0 |
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