Question

If two circles of radii $5$ units touches each other at $(1,2)$ and the equation of the common tangent is $4x+3y=10$, then the equation of the circle is/are

• A. $x^2+y^2-10x-10y+25=0$
• B. $x^2+y^2+6x+2y-15=0$
• C. $x^2+y^2-10x-10y-25=0$
• D. $x^2+y^2+6x-2y+15=0$

Answer 1

Answer: (A), (B)

$x^2+y^2+6x+2y-15=0$

Let the centre of the circle be $(h,k)$
Given tangent is $4x+3y=10$
It passes through $(1,2)$
Line passing through $(1,2)$ and centre is perpendicular to the given tangent, so
$\frac{k-2}{h-1}=\frac{3}{4}\cdots \left(1\right)$

Now, the radius of circle
$\sqrt{(h-1{)}^2+(k-2{)}^2}=5\Rightarrow (h-1{)}^2+(k-2{)}^2=25$
Using equation $\left(1\right)$, we get
$\Rightarrow (h-1{)}^2+\frac{9}{16}(h-1{)}^2=25\Rightarrow (h-1{)}^2=16\Rightarrow h-1=\pm 4\Rightarrow h=5,-3\Rightarrow k=5,-1$

Hence, the circles are
$x^2+y^2-10x-10y+25=0$ and
$x^2+y^2+6x+2y-15=0$