# Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes

**Solution:**

Let the equation of the required circle be (x - h)^{2} + (y - k)^{2} = r^{2}

Since the circle passes through (0, 0) ,

(0 - h)^{2} + (0 - k)^{2} = r^{2}

⇒ h^{2} + k^{2} = r^{2}

The equation of the circle now becomes (x - h)^{2} + (y - k)^{2} = h^{2} + k^{2}

It is given that the circle makes intercepts a and b on the coordinate axes.

This means that the circle passes through points (a, 0) and (0, b)

Therefore,

(a - h)^{2} + (0 - k)^{2} = h^{2} + k^{2}

(0 - h)^{2} + (b - k)^{2} = h^{2} + k^{2}

From equation (1) , we obtain

⇒ a^{2} + h^{2} - 2ah + k^{2} = h^{2} + k^{2}

⇒ a^{2} - 2ah = 0

⇒ a (a - 2h) = 0

⇒ a = 0 or (a - 2h) = 0

However, a ≠ 0

Hence,

(a - 2h) = 0

⇒ h = a/2

From equation (2) , we obtain

h^{2} + b^{2} - 2bk + k^{2} = h^{2} + k^{2}

⇒ b^{2} - 2bk = 0

⇒ b (b - 2k) = 0

⇒ b = 0 or (b - 2k) = 0

However, b ≠ 0

Hence,

(b - 2k ) = 0

⇒ k = b/2

Thus, the equation of the required circle is

(x - a/2)^{2} + (y - b/2)^{2} = (a/2)^{2} + (b/2)^{2}

[(2x - a)/2]^{2} + [(2y - b)/2]^{2} = (a^{2} + b^{2})/4

⇒ 4x^{2} - 4ax + a^{2} + 4y^{2} - 4by + b^{2} = a^{2} + b^{2}

⇒ 4x^{2} + 4y^{2} - 4ax - 4by = 0

⇒ 4 (x^{2} + y^{2} - ax - by) = 0

⇒ x^{2} + y^{2} - ax - by = 0

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 Question 13

## Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes

**Summary:**

The equation of the circle is x^{2} + y^{2} - ax - by = 0, passing through points (a, 0) and (0, b)

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