# 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.