RD Sharma Solutions Class 11 The Circle

RD Sharma Solutions Class 11 Chapter 24

A circle is defined as the locus of a point that moves in a plane such that its distance from any fixed point in that plane is always same. This fixed point is called the centre of the circle and the constant distance is called the radius of the circle.

Equation of a Circle:

\((x-h)^{2}\;+\;(y-k)^{2}\;=\;a^{2}\)

Here, (h, k) = The coordinates of the centre of circle, a = radius of circle and (x, y) = coordinates of ant point on the circumference of a circle.

Case – 1: When the centre of circle coincides with the origin:

Here, k = h = 0. Therefore, the equation of circle becomes: \((x)^{2}\;+\;(y)^{2}\;=\;a^{2}\)

Case – 2: When the circle touches y axis:

Here, h = a (Radius of circle). Therefore, the equation of circle becomes: \((x-a)^{2}\;+\;(y-k)^{2}\;=\;a^{2}\)

i.e. \((x)^{2}\;+\;(y)^{2}\;-\;2ax\;-\;2ky\;+\;k^{2}\) = 0

Case – 3: When the circle touches x axis:

Here, k = a (Radius of circle). Therefore, the equation of circle becomes: \((x-h)^{2}\;+\;(y-a)^{2}\;=\;a^{2}\)

i.e. \((x)^{2}\;+\;(y)^{2}\;-\;2ay\;-\;2hx\;+\;h^{2}\) = 0

Case – 4: When the circle passes through the origin and the centre lies on the x axis:

Here, k = 0 and h = a (Radius of circle). Therefore, the equation of circle becomes: \((x-a)^{2}\;+\;(y-0)^{2}\;=\;a^{2}\)

i.e. \((x)^{2}\;+\;(y)^{2}\;-\;2ax\) = 0

Case – 5: When the circle passes through the origin and the centre lies on the y axis:

Here, h = 0 and k = a (Radius of circle). Therefore, the equation of circle becomes: \((x-0)^{2}\;+\;(y-a)^{2}\;=\;a^{2}\)

i.e. \((x)^{2}\;+\;(y)^{2}\;-\;2ay\) = 0

RD Sharma Solutions for Class 11 Circles will help you to solve different problems related to circles like, finding the radius and centre of any given circle, finding the equation of a circle when its centre and radius are known, finding the equation of circle passing through three given points and more in a very simplified and organized way. Practice exercise wise solved RD Sharma Solutions by clicking the links given below that will help you to answer certain tricky questions asked in competitive exams like JEE Mains and JEE Advanced accurately.