Question

Equation of circle touching x axis and y axis and perpendicular distance of centre of circle from 3x+4y+11=0

Answer 1

touching X axis =-11/3
y axis =-11/4
perpendicular distance from centre=3x+4y+11/root(9+16)=3x+4y+11/5
if centre is origin then it becomes 11/5

Answer 2

Answer is $x^2+y^2+4x+4y+4=0$

Step-by-step explanation:

• 3x+4y+11=0 is the equation of the circle

Let ′a′ be the radius of the circle.

Hence, Center of the circle is (-a,-a).

• The line which touches the circle is 3x−4y+8=0

Clearly this is a tangent to the circle.

∴ The perpendicular distance = a units

i.e  

$\left | \frac {3(-a)-4(-a)+8}{\sqrt {3^2 +4^2}} \right |$

$a = \frac {-3a+4a+8}{5}$

So, 5a = a+8

or a = 2

Hence, Equation of required circle is $[(x-(-2))]^2 + [(y-(-2))]^2 = 2^2$

i.e. $(x+2)^2 + (y+2)^2 = 4$

On expanding we get -

$x^2 + 4x + 4 + y^2 + 4y + 4 = 4$

or $x^2 +y^2 + 4x + 4y + 4 = 0$ is the desired equation of required circle.