Circle Examples

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

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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

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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.