Q. 64.3( 7 Votes )

# Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0and lies in the third quadrant.

Answer :

As the circle lies in third quadrant, then the centre is (-a, -a).

Perpendicular Distance (Between a point and line) = , whereas the point is and the line is expressed as ax + by + c = 0

The line which touches the circle is 3x−4y+8=0, which is a tangent to the circle.

∴ The perpendicular distance = a units (radius of the circle)

Co-ordinates of the centre of the circle = (-2,-2)

Since, the equation of a circle having centre (h,k), having radius as "r" units, is

(x – h)^{2} + (y – k)^{2} = r^{2}

(x – (-2))^{2} + (y – (-2))^{2} = 2^{2}

(x + 2)^{2} + (y + 2)^{2} = 4

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

x^{2} + y^{2} + 4x + 4y + 4 = 0

The equation of the given circle is x^{2} + y^{2} + 4x + 4y + 4 = 0.

Rate this question :

The equation of the circle which touches the axes of coordinates and the line and whose centres lie in the first quadrant is x^{2} + y^{2} – 2cx – 2cy + c^{2} = 0, where c is equal to

Find the equation of the circle which circumscribes the triangle formed by the lines:

y = x + 2, 3y = 4x and 2y = 3x

RD Sharma - MathematicsFind the equation of the circle which passes through (3, - 2), (- 2, 0) and has its centre on the line 2x – y = 3.

RD Sharma - MathematicsShow that the points (5, 5), (6, 4), (- 2, 4) and (7, 1) all lie on a circle, and find its equation, centre, and radius.

RD Sharma - MathematicsIf the circle x^{2} + y^{2} + 2ax + 8y + 16 = 0 touches x - axis, then the value of a is

Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on line x – 4y = 1.

RD Sharma - MathematicsFind the equation of the circle passing through the points :

(5, - 8), (- 2, 9) and (2, 1)

RD Sharma - MathematicsThe circle x^{2} + y^{2} + 2gx + 2 fy + c = 0 does not intersect x - axis, if