Equations of Concentric Circles

We will learn how to form the equation of concentric circles.

Two circles or more than that are said to be concentric if they have the same centre but different radii.

Let, x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 be a given circle having centre at (- g, - f) and radius = \(\mathrm{\sqrt{g^{2} + f^{2} - c}}\).

Therefore, the equation of a circle concentric with the given circle x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c = 0 is

x\(^{2}\) + y\(^{2}\) + 2gx + 2fy + c' = 0 

Both the circle have the same centre (- g, - f) but their radii are not equal (since, c ≠ c')

Similarly, the equation of a circle with centre at (h, k) and radius equal to r, is (x - h)\(^{2}\) + (y - k)\(^{2}\) = r\(^{2}\).

Therefore, the equation of a circle concentric with the circle (x - h)\(^{2}\) + (y - k)\(^{2}\) = r\(^{2}\) is  (x - h)\(^{2}\) + (y - k)\(^{2}\) = r\(_{1}\)\(^{2}\), (r\(_{1}\) ≠ r)      

Assigning different values to r\(_{1}\) we shall have a family of circles each of which is concentric with the circle (x - h)\(^{2}\) + (y - k)\(^{2}\) = r\(^{2}\).


Solved example to find the equation of a concentric circle:

Find the equation of the circle which is concentric with the circle 2x\(^{2}\) + 2y\(^{2}\) + 3x - 4y + 5 = 0 and whose radius is 2√5 units.

Solution:         

2x\(^{2}\) + 2y\(^{2}\) + 3x - 4y + 5 = 0

⇒ x\(^{2}\) + y\(^{2}\) + 3/2x - 2y + \(\frac{5}{2}\) = 0 ………………..(i)

Clearly, the equation of a circle concentric with the circle (i) is

x\(^{2}\) + y\(^{2}\) + \(\frac{3}{2}\)x - 2y + c = 0 ……………………..(ii)

Now, the radius of the circle (ii) = \(\sqrt{(\frac{3}{2})^{2} + (-2)^{2} - c}\)

By question, \(\sqrt{\frac{9}{4} + 4 - c}\) = 2√5

⇒ \(\frac{25}{4}\) - c = 20

⇒ c = \(\frac{25}{4}\) - 20

c = -\(\frac{55}{4}\)

Therefore, the equation of the required circle is

x\(^{2}\) + y\(^{2}\) + \(\frac{3}{2}\)x - 2y - \(\frac{55}{4}\) = 0

⇒ 4x\(^{2}\) + 4y\(^{2}\) + 6x - 8y - 55 = 0.

 The Circle




11 and 12 Grade Math 

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