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Worked Examples · Example 12

Q.Find the equation of the circle of radius 5 whose centre lies on x-axis and passes through the point (2,3)(2, 3).

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Let the centre be (a,0)(a,0) on the x-axis; use the distance to (2,3)(2,3) equal to the radius 5 to find aa, giving two possible circles.

Circle with centre (h,k)(h,k), radius rr: (x−h)2+(y−k)2=r2(x-h)^2+(y-k)^2=r^2. Distance formula used to fix the unknown centre.

  1. Let the centre be (a,0)(a,0) since it lies on the x-axis, with radius r=5r=5.
  2. The circle passes through (2,3)(2,3), so:

(a−2)2+(3−0)2=52(a-2)^2+(3-0)^2=5^2

(a−2)2+9=25⇒(a−2)2=16(a-2)^2+9=25\Rightarrow(a-2)^2=16

  1. Take square roots: a−2=±4⇒a=6 or a=−2a-2=\pm4\Rightarrow a=6\ \text{or}\ a=-2.
  2. Case a=6a=6: centre (6,0)(6,0): (x−6)2+y2=25⇒x2−12x+36+y2=25⇒x2+y2−12x+11=0(x-6)^2+y^2=25\Rightarrow x^2-12x+36+y^2=25\Rightarrow x^2+y^2-12x+11=0. …

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