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

**Solution:**

The equation of the given circle is 2x^{2} + 2y^{2} - x = 0

⇒ 2x^{2} + 2y^{2} - x = 0

⇒ (2x^{2} - x) + 2y^{2} = 0

⇒ 2 [(x^{2}- x/2) + y^{2}] = 0

⇒ {x^{2} - 2 (x)(1/4) + (1/4)^{2}} + y^{2} - (1/4)^{2} = 0

⇒ (x - 1/4)^{2} + (y - 0)^{2} = (1/4)^{2}

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

Therefore, on comparing both equations we get

h = 1/4, k = 0 and r = 1/4

Thus, the center of the given circle is (1/4, 0) while its radius is 1/4

NCERT Solutions Class 11 Maths Chapter 11 Exercise 11.1 Question 9

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

**Summary:**

The center and radius of the circle are (1/4, 0) and 1/4 respectively