Equation of the circle having its centre on the line $2 x+y$ $+3=0$ and having the lines $3 x+4 y-18=0,3 x+4…
Equation of the circle having its centre on the line $2 x+y$ $+3=0$ and having the lines $3 x+4 y-18=0,3 x+4 y+2=$ 0 as tangents is
$x^2+y^2+6 x+8 y+4=0$
$x^2+y^2-6 x-8 y+18=0$
$x^2+y^2-8 x+10 y+37=0$
$x^2+y^2+8 x-10 y+37=0$
Solution
Equations of tangents
$\Rightarrow 3 x+4 y-18=0 \Rightarrow 3 x+4 y+2=0$
Since slope of tangents are equal. So tangents are parallel. And distance between two parallel lines is
$d=\frac{\left|c_2-c_1\right|}{\sqrt{a^2+b^2}}=\frac{20}{5}=4 \Rightarrow$ Radius $=\frac{4}{2}=2$
Let $(h, k)$ be the centre and radius is perpendicular on tangent
$\therefore$ perpendicular distance from centre to tangent $=$ radius
$\Rightarrow \frac{3 h+4 k+2}{5}=2 \Rightarrow 3 h+4 k-8=0$ ....(i)
Also $(h, k)$ lies on $2 x+y+3=0$
$\Rightarrow 2 h+k+3=0$ ...(ii)
From (i) and (ii), $h=-4, k=5$
$\therefore$ Equation of circle is $x^2+y^2+8 x-10 y+37=0$