Question

Two circles, each of radius 5 units, touch each other at (1, 2). If the equation of their common tangent is 4x + 3y – 10 = 0 then equation of one such circle is

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

Answer 1

The correct option is B

$x^2+y^2–10x–10y+25=0$

Let the point of contact be $(x_1,y_1)$ = (1, 2)
We know that co-ordinates of any point on the circle can be given by -
$a\pm rcos\theta ,b\pm rsin\theta )$
Where (a,b) are the co-ordinates of the center of the circle and $\theta )$ is the angle made by the line joining that point and the center with the horizontal axis(X-axis).
We are the given the equation of the tangent.
We know that the radius is perpendicular to the tangent. So the slope of radius will be negative inverse of the slope of the tangent.
Slope of the tangent is $\frac{-4}{3}$
And slope of the radius will be - $\frac{3}{4}$
This slope is nothing but the tan of the angle it makes with the horizontal.
So, tan($\theta$) = $\frac{3}{4}$
$cos\theta =\frac{4}{5},sin\theta =\frac{3}{5}$
Now we can write -
(1,2 ) = $a\pm 5cos\theta ,b\pm 5sin\theta )$
(1,2 ) = $a\pm 4,b\pm 3)$
On comparing we get -
$\Rightarrow centre=(1\pm 4,2\pm 3)=(5,5)or(-3,-1)$
Using this we can write the equation of circle.
$x^2+y^2–10x–10y+25=0$
$x^2+y^2+6x+2y–15=0$