Question

Find the equation of the circle which has centre C (3,1) and which touches the line 5x-12y + 10 = 0.​

Answer 1

Answer:

Step-by-step explanation:

Centre of the circle is $\,\,\rm{\equiv(3,1)}$

Since $\,\,\rm{5{x}-12{y}+10=0}\,\,$ is the tangent to the circle, so, radius will be the perpendicular distance from center.

We know,

$\boxed{\bf{Perpendicular\,\,distance\,\,from\,\,a\,\,point\,\,is\,\,given\,\,by\,\,,\,p=\dfrac{\left|a{x}_{1}+b{y}_{1}+c\right|}{{a}^{2}+{b}^{2}}}}$

So,

$\rm{radius\,,\,\,r=\dfrac{\left|5(3)-12(1)+10\right|}{\left(5\right)^2+\left(-12\right)^2}}$

$\rm{\implies\,r=\dfrac{\left|15-12+10\right|}{25+144}}$

$\rm{\implies\,r=\dfrac{13}{169}}$

$\rm{\implies\,r=\dfrac{1}{13}}$

Now, the equation of the circle is

$\sf{\left(y-1\right)^{2}+\left(x-3\right)^{2}=\left(\dfrac{1}{13}\right)^2}$

$\sf{\implies\,{y}^{2}-2{y}+1+{x}^{2}-6{x}+9=\dfrac{1}{169}}$

$\sf{\implies\,{x}^{2}+{y}^{2}-6{x}-2{y}+10=\dfrac{1}{169}}$

$\sf{\implies\,169\,{x}^{2}+169\,{y}^{2}-1014\,x-338\,y+1690=1}$

$\sf{\implies\,169\,{x}^{2}+169\,{y}^{2}-1014\,x-338\,y+1689=0}$