Question

Two circles with radius $5$ touches at the point $(1,2)$. If the equation of common tangent is $4x+3y=10$ and one of the circle is $x^2+y^2+6x+2y-15=0$. Find the equation of other circle.

• A. ${(x-5)}^2+{(y+5)}^2=25$
• B. ${(x+5)}^2+{(y-5)}^2=25$
• C. ${(x-5)}^2+{(y-5)}^2=25$
• D. ${(x+5)}^2+{(y+5)}^2=25$

Answer 1

Answer: (C)

${(x-5)}^2+{(y-5)}^2=25$

Let the centre of the unknown circle be $C=(x,y)$.
The centre of the other circle is $(-3,-1)$.
The line joining the centres of the circle is perpendicular to the common tangent.
If the slope of the tangent is $\frac{-4}{3}$
Then slope of the line joining the centres is $\frac{3}{4}$.
Thus equation of the line
$\frac{y+1}{x+3}=\frac{3}{4}$
$4y+4=3x+9$
$y=\frac{3x+5}{4}$ ...(i)
Hence the centre is of the form
$C=(x,\frac{3x+5}{4})$.
Since the radius of the circle is 5 units .
Hence $CP=5units$ where $P=(1,2)$.
$C{P}^2=25$
$(x-1{)}^2+(\frac{3x+5}{4}-2{)}^2=25$
$(x-1{)}^2+(\frac{3x-3}{4}{)}^2=25$
$(x-1{)}^2[1+\frac{9}{16}]=25$
$(x-1{)}^2(\frac{25}{16})=25$
$(x-1{)}^2=16$
$x-1=\pm 4$
$x=-3$ and $x=5$
$\Rightarrow y=-1$ and $y=5$
Since $(-3,-1)$ is the centre of the other circle, hence the centre of the required circle is
$C=(5,5)$.
Thus equation of the circle is
$(x-5{)}^2+(y-5{)}^2=25$