Question

find the equation of circle with radius 1/12 and centre (1/2 , 1/4)​

Answer 1

ANSWER:-

Given:

•Radius of the circle,r= 1/12.

•Centre of the circle = (1/2, 1/4).

To find:

The equation of the circle.

Solution:

We know that equation of a circle;

=) (x-h)² + (y-k)²= r²

⚫It's given that centre (h,k)=(1/2, 1/4).

&

⚫Radius,(r)=(1/12).

Therefore,

Putting the given value of the equation of the circle:

$(x - \frac{1}{2} ) {}^{2} + (y - \frac{1}{4} ) {}^{2} = ( \frac{1}{12} ) {}^{2} \\ \\ = > {x}^{2} + ( { \frac{1}{2} )}^{2} - 2 \times x \times \frac{1}{2} + {y}^{2} + ( { \frac{1}{4}) }^{2} - 2 \times y \times \frac{1}{4} = \frac{1}{144} \\ \\ = > {x}^{2} + \frac{1}{4} - x + {y}^{2} + \frac{1}{16} - \frac{y}{2} = \frac{1}{144} \\ \\ = > {x}^{2} + {y}^{2} - x - \frac{y}{2} + \frac{1}{4} + \frac{1}{16} - \frac{1}{144} = 0 \\ \\ = > \frac{144 {x}^{2} + 144 {y}^{2} - 144x - 72y + 34 + 9 + 1}{144} = 0 \\ \\ = > 144 {x}^{2} + 144 {y}^{2} - 144x - 72y + 44 = 0 \\ \\ = > 4(36 {x}^{2} + 36 {y}^{2} - 36x - 18y + 11) = 0 \\ \\ = > 36 {x}^{2} + 36 {y}^{2} - 36x - 18y + 11 = 0$

Thus,

The equation of the circle is;

=) 36x² + 36y² - 36x - 18y + 11=0.

Hope it helps ☺️