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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equation of the circle passing through the point $(2,3)$ with radius $5$ whose centre lies on the x-axis.

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Toolbox:
  • Given C (h, k) be the centre and r the radius of circle. Let P(x, y) be any point on the circle. Then, by the definition, | CP | = r . By the distance formula, we have, $(x-h)^2+(y-k)^2=r^2$
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  • If we are given the radius and one of the intersection coordinates of a circle, we can substitute those values in the equation of the circle and solve for the centre.
If we are given the radius and one of the intersection coordinates of a circle, we can substitute those values in the equation of the circle and solve for the centre.
Given that the centre lies on the x-axis $\rightarrow k = 0$
Given that the circle has a radius $r=5$ and passes through $(2,3) \rightarrow (2-h)^2+(3-0)^2 = 5^2 \rightarrow (2-h)^2 = 25-9 = 16 $
$\Rightarrow (2-h) = \pm 4 $
$\Rightarrow$ If $2-h = 4 \rightarrow h = -2$ and if $2-h = -4 \rightarrow h = 6$
Case-1, when $h=-2$:
The Equation of the circle = $(x+2)^2 + y^2 = 25 \rightarrow x^2 + y^2 + 4x - 21 = 0$
Case-2, when $h=6$:
The Equation of the circle = $(x-6)^2 + y^2 = 25 \rightarrow x^2 + y^2 -12x +11 = 0$
answered Apr 1, 2014 by balaji.thirumalai
edited Apr 1, 2014 by balaji.thirumalai
 

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