Question

Find equation of circle which touches The line x=0 y=0 and 3x + 4y=4

Answer 1

Solution:

Keep this statement in mind: Length of tangents from external point to a circle are equal.

OT=OM=c

BM=B N=a

AT=AN=b

The three line which touches the circle are:

x=0 ,y=0 and 3 x + 4 y=4

Plotting the lines on the coordinate plane and drawing the circle which touches these three lines

1. a +c=1

2. b+c=$\frac{4}{3}$

3. a+b= $\sqrt{[\frac{4}{3}]^2+1^2]}=\frac{5}{3}$

(1) - (2)

4. a - b= $\frac{-1}{3}$

(3) + (4)

2 a = $\frac{4}{3}$

a= $\frac{2}{3}$

Putting the value of a in equation- 1

c= 1 -$\frac{2}{3}$

c= $\frac{1}{3}$

Putting the value of a in equation- 4

$\frac{2}{3}$ - b= $\frac{-1}{3}$

b=1

Center of circle will be (  $\frac{1}{3}$,  $\frac{1}{3}$  )

And , Radius =  $\frac{1}{3}$

Equation of circle having Center (  $\frac{1}{3}$,  $\frac{1}{3}$  ) and Radius  $\frac{1}{3}$ is

$(x- \frac{1}{3})^2 +(y- \frac{1}{3})^2=[\frac{1}{3}]^2\\\\ x^2 + y^2-\frac{2 x}{3}-\frac{2 y}{3}+\frac{1}{9}=0$