Circle Examples

Here you will learn some circle examples for better understanding of circle concepts.

Example 1 : Find the equation of the normal to the circle \(x^2 + y^2 – 5x + 2y -48\) = 0 at the point (5,6).

Solution : Since the normal to the circle always passes through the centre so the equation of the normal will be the line passing through (5,6) & (\(5\over 2\), -1)

i.e.   y + 1 = \(7\over {5/2}\)(x – \(5\over 2\)) \(\implies\) 5y + 5 = 14x – 35

\(\implies\)   14x – 5y – 40 = 0



Example 2 : Find the equation of circle having the lines \(x^2\) + 2xy + 3x + 6y = 0 as its normal and having size just sufficient to contain the circle x(x – 4) + y(y – 3) = 0.

Solution : Pair of normals are (x + 2y)(x + 3) = 0

\(\therefore\)   Normals are x + 2y = 0, x + 3 = 0

Point of intersection of normals is the centre of the required circle i.e. \(C_1\)(-3,3/2) and centre of the given circle is \(C_2\)(2,3/2) and radius \(r^2\) = \(\sqrt{4 + {9\over 4}}\) = \(5\over 2\)

Let \(r_1\) be the radius of the required circle

\(\implies\)   \(r_1\) = \(C_1\)\(C_2\) + \(r_2\) = \(\sqrt{(-3-2)^2 + ({3\over 2}- {3\over 2})^2}\) + \(5\over 2\) = \(15\over 2\)

Hence equation of required circle is \(x^2 + y^2 + 6x – 3y – 45\) = 0



Example 3 : The equation of the circle through the points of intersection of \(x^2 + y^2 – 1\) = 0, \(x^2 + y^2 – 2x – 4y + 1\) = 0 and touching the line x + 2y = 0, is-

Solution : Family of circles is \(x^2 + y^2 – 2x – 4y + 1\) + \(\lambda\)(\(x^2 + y^2 – 1\)) = 0

(1 + \(\lambda\))\(x^2\) + (1 + \(\lambda\))\(y^2\) – 2x – 4y + (1 – \(\lambda\))) = 0

\(x^2 + y^2 – {2\over {1 + \lambda}} x – {4\over {1 + \lambda}}y + {{1 – \lambda}\over {1 + \lambda}}\) = 0

Centre is (\({1\over {1 + \lambda}}\), \({2\over {1 + \lambda}}\))   and radius = \(\sqrt{4 + {\lambda}^2}\over |1 + \lambda|\)

Since it touches the line x + 2y = 0, hence

Radius = Perpendicular distance from centre to the line

i.e., |\({1\over {1 + \lambda}} + 2{2\over {1 + \lambda}}\over \sqrt{1^2 + 2^2}\)| = \(\sqrt{4 + {\lambda}^2}\over |1 + \lambda|\) \(\implies\) \(\sqrt{5}\) = \(\sqrt{4 + {\lambda}^2}\) \(\implies\) \(\lambda\) = \(\pm\) 1.

\(\lambda\) = -1 cannot be possible in case of circle. So \(\lambda\) = 1.

Hence the equation of the circle is \(x^2 + y^2 – x – 2y\) = 0


Practice these given circle examples to test your knowledge on concepts of circles.

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