Question

The equation of the circle which touches both the axes and the straight line 4x + 3y = 6 in the first quadrant and lies below it is :

• A. 4x² + 4y²– 4x – 4y + 1 = 0
• B. 4x² + y²– 6x – 6y + 9 = 0
• C. x² + y²– 6x – y + 9 = 0
• D. 4(x² + y²– x – 6y) + 1 = 0

Answer 1

Answer: (A)

4x² + 4y²– 4x – 4y + 1 = 0

Answer 2

Let radius of circle is r

Ž C ŗ (r, r) Q Pt. touches 4x + 3y – 6 = 0

P = r Ž $\frac{|4r + 3r - 6|}{5} = r$ Ž 7r – 6 = ± 5r

12r = 6, 2r = 6 Ž r = 1/2, r = 3

Q r = 3 not possible

Ž r = 1/2 Equation (x –1/2)² + (y –1/2)² = ¼