We will learn how to find the equation of a circle through the intersection of two given circles.

The equation of a family of circles passing through the intersection of the circles P\(_{1}\) = x\(^{2}\) + y\(^{2}\) + 2g\(_{1}\)x + 2f\(_{1}\)y + c\(_{1}\) = 0 and P\(_{2}\) = x\(^{2}\) + y\(^{2}\) + 2g\(_{2}\)x + 2f\(_{2}\)y + c\(_{2}\) = 0 is P\(_{1}\) + λP\(_{2}\) = 0 i.e., (x\(^{2}\) + y\(^{2}\) + 2gx\(_{1}\) + 2fy\(_{1}\) + c\(_{1}\)) + λ(x\(^{2}\) + y\(^{2}\) + 2g\(_{2}\)x + 2f\(_{2}\)y + c\(_{2}\)) = 0, where λ (≠ -1) in an arbitrary real number.

**Proof:**

Let the equations of the given circles be

P\(_{1}\) = x\(^{2}\) + y\(^{2}\) + 2g\(_{1}\)x + 2f\(_{1}\)y + c\(_{1}\) = 0 ………………………..(i) and

P\(_{2}\) = x\(^{2}\) + y\(^{2}\) + 2g\(_{2}\)x + 2f\(_{2}\)y + c\(_{2}\) ………………………..(ii)

Consider the equation P\(_{1}\) + λP\(_{2}\) = 0 i.e., the equation of any curve through the points of intersection of the circles (1) and (2) is

(x\(^{2}\) + y\(^{2}\) + 2g\(_{1}\)x + 2f\(_{1}\)y + c\(_{1}\)) + λ(x\(^{2}\) + y\(^{2}\) + 2g\(_{2}\)x + 2f\(_{2}\)y + c\(_{2}\)) = 0 ………………………..(iii)

Clearly, it represents a circle for all values of λ except λ = -1. For λ = -1 (iii) becomes a first degree equation in x, y which represents a line. In order to prove that it passes through the points of intersection of the two given circles, it is sufficient to show that their points of intersection satisfy (iii).

Let (x\(_{1}\), y\(_{1}\)) be a point of intersection of the given circles.

Then,

\(\mathrm{x_{1}^{2} + y_{1}^{2} + 2g_{1}x_{1} + 2f_{1}y_{1} + c_{1}}\) and \(\mathrm{x_{1}^{2} + y_{1}^{2} + 2g_{2}x_{1} + 2f_{2}y_{1} + c_{2}}\)

⇒ (\(\mathrm{x_{1}^{2} + y_{1}^{2} + 2g_{1}x_{1} + 2f_{1}y_{1} + c_{1}}\)) + λ(\(\mathrm{x_{1}^{2} + y_{1}^{2} + 2g_{2}x_{1} + 2f_{2}y_{1} + c_{2}}\)) = 0 + λ0 = 0

⇒ (x\(_{1}\), y\(_{1}\)) lies on (iii).

Similarly, it can be proved that the second point of intersection of the given circles also satisfy (i)

Hence, (iii) gives the family of circles passing through the intersection of the given circles.

In other words, the equation of any curve through the points of intersection of the circles (i) and (ii) is

(x\(^{2}\) + y\(^{2}\) + 2g\(_{1}\)x + 2f\(_{1}\)y + c\(_{1}\)) +
λ(x\(^{2}\) + y\(^{2}\) + 2g\(_{2}\)x + 2f\(_{2}\)y + c\(_{2}\))………………………..(iv)

⇒ (1 + λ)(x\(^{2}\) + y\(^{2}\)) + 2(g\(_{1}\) + g\(_{2}\)λ)x + 2(f\(_{1}\) + f\(_{2}\)λ)y + c\(_{1}\) + λc\(_{2}\) = 0

⇒ x\(^{2}\) + y\(^{2}\) + 2 ∙ \(\mathrm{\frac{g_{1} + g_{2}λ}{1 + λ}}\) x + 2 ∙ \(\mathrm{\frac{f_{1} + f_{2}λ}{1 + λ}}\)y + \(\mathrm{\frac{c_{1} + c_{2}λ}{1 + λ}}\) = 0 ………………………..(v)

If λ ≠ - 1, then equation (v) will represent the equation of a circle. Therefore, the equation (iv) represents the family of circles through the points of intersection of the circles (1) and (2).

Solved examples to find the equations of a circle through the points of intersection of two given circles:

**1.** Find the equation of the circle through the intersection of the circles x\(^{2}\) + y\(^{2}\) - 8x - 2y + 7 = 0 and x\(^{2}\) + y\(^{2}\) - 4x + 10y + 8 = 0 and passes through the point (-1, -2).

**Solution:**

The equation of any circles passing through the intersection of the circles S\(_{1}\) = x\(^{2}\) + y\(^{2}\) - 8x - 2y + 7 = 0 and S\(_{2}\) = x\(^{2}\) + y\(^{2}\) - 4x + 10y + 8 = 0 is S\(_{1}\) + λS\(_{2}\) = 0

Therefore, the equation of the required circle is (x\(^{2}\) + y\(^{2}\) - 8x - 2y + 7) + λ(x\(^{2}\) + y\(^{2}\) - 4x + 10y + 8) = 0, where λ (≠ -1) in an arbitrary real number

This circle passes through the point (-1, -2), therefore,

(1 + λ) + 4(1 + λ) + 4(2 + λ) + 4(1 - 5λ) + 7 + 8λ = 0

⇒ 24 - 3λ = 0

⇒ λ = 8

Now putting the value of λ = 8 in the equation (x\(^{2}\) + y\(^{2}\) - 8x - 2y + 7) + λ(x\(^{2}\) + y\(^{2}\) - 4x + 10y + 8) = 0 we get the required equation as 9x\(^{2}\) + 9y\(^{2}\) – 40x + 78y + 71 = 0.

**2.** Find the equation of the circle through the intersection of the circles x\(^{2}\) + y\(^{2}\) - x + 7y - 3 = 0 and x\(^{2}\) + y\(^{2}\) - 5x - y + 1 = 0, having its centre on the line x + y = 0.

**Solution:**

x\(^{2}\) + y\(^{2}\) - x + 7y - 3 + λ(x\(^{2}\) + y\(^{2}\) - 5x - y + 1) = 0, (λ ≠1)

⇒(1 + λ) (x\(^{2}\) + y\(^{2}\)) - (1 + 5λ)x + (7 - λ)y - 3 + λ = 0

⇒ x\(^{2}\) + y\(^{2}\) - \(\frac{1 + 5λ}{1 + λ}\)x - \(\frac{λ - 7}{1 + λ}\)y + \(\frac{λ - 3}{1 + λ}\) = 0 …………….(i)

Clearly, the co-ordinates of the centre of the circle (i) are [\(\frac{1 + 5λ}{2(1 + λ)}\), \(\frac{λ - 7}{2(1 + λ)}\)] By question, this point lies on the line x + y = 0.

Therefore, \(\frac{1 + 5λ}{2(1 + λ)}\) + \(\frac{λ - 7}{2(1 + λ)}\) = 0

⇒1 + 5λ + λ - 7 = 0

⇒ 6λ = 6

⇒ λ = 1

Therefore, the equation of the required circle is 2(x\(^{2}\) + y\(^{2}\)) - 6x + 6y - 2 = 0, [putting λ = 1 in (1)]

⇒ x\(^{2}\) + y\(^{2}\) - 3x + 3y - 1 = 0.

**●** **The Circle**

**Definition of Circle****Equation of a Circle****General Form of the Equation of a Circle****General Equation of Second Degree Represents a Circle****Centre of the Circle Coincides with the Origin****Circle Passes through the Origin****Circle Touches x-axis****Circle Touches y-axis****Circle Touches both x-axis and y-axis****Centre of the Circle on x-axis****Centre of the Circle on y-axis****Circle Passes through the Origin and Centre Lies on x-axis****Circle Passes through the Origin and Centre Lies on y-axis****Equation of a Circle when Line Segment Joining Two Given Points is a Diameter****Equations of Concentric Circles****Circle Passing Through Three Given Points****Circle Through the Intersection of Two Circles****Equation of the Common Chord of Two Circles****Position of a Point with Respect to a Circle****Intercepts on the Axes made by a Circle****Circle Formulae****Problems on Circle**

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