Question

The equation of the circle which touches the lines $3x+4y-5=0$ and $3x+4y+25=0$ and whose centre lies on the line $x+2y=0$, is

• A. $(x+10{)}^2+(y-5{)}^2=16$
• B. $(x+5{)}^2+(y-9{)}^2=9$
• C. $(x+10{)}^2+(y-5{)}^2=9$
• D. $(x+5{)}^2+(y-9{)}^2=9$

Answer 1

The correct option is C $(x+10{)}^2+(y-5{)}^2=9$

The lines $3x+4y-5=0$ and $3x+4y+25=0$ are parallel to each other and are touching the circle.


Let $C$ be the centre of the circle.
Let the line $x+2y=0$ intersects the given parallel lines at points $A$ and $B$ respectively.
Coordinates of $A$ is $(5,-2.5)$ and coordinates of $B$ is $(-25,12.5)$.
The centre of the circle $C$ is the mid-point of $AB$.
$\therefore$ Coordinates of $C$ are $(\frac{5-25}{2},\frac{-2.5+12.5}{2})$
i.e., coordinates of $C$ are $(-10,5)$
Radius $=\frac{1}{2}PQ=\frac{1}{2}\left|\frac{-5-25}{\sqrt{3^2+4^2}}\right|=3$

Hence, equation of the circle is $(x+10{)}^2+(y-5{)}^2=9$.