# Find the centre and radius of the circle x^{2} + y^{2} - 4x - 8y - 45 = 0

**Solution:**

The equation of the given circle is

x^{2} + y^{2} - 4x - 8y - 45 = 0

⇒ x^{2} + y^{2} - 4x - 8y - 45 = 0

⇒ (x^{2} - 4x) + (y^{2} - 8y) = 45

⇒ {x^{2} - 2(x)(2) + 2^{2}} + {y^{2} - 2 (y )(4) + 4^{2}} - 4 - 16 = 45

⇒ ( x - 2)^{2} + (y - 4)^{2} = 65

⇒ ( x - 2)^{2} + ( y - 4)^{2} = (√65)^{2}

which is of the form (x - h)^{2} + (y - k)^{2} = r^{2}

Therefore, on comparing both equations we get

h = 2, k = 4 and r = √65

Thus, the centre of the given circle is (2, 4) while its radius is √ 65

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 Question 7

## Find the centre and radius of the circle x^{2} + y^{2} - 4x - 8y - 45 = 0

**Summary:**

The center and radius of the circle are (2, 4) and √ 65 respectively

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