Question

Find the equation of the circle whose centre is (3, -1) and which cut-off an intercept of length 6 from the line $2x-5y+18=0$.

Answer 1

The centre of the circle is 0(3, -1)
The circle cuts-off an intercept of length 6 unit from $2x-5y+18=0\dots \left(1\right)$
Let, AB = 6
$\Rightarrow$ AM = 3 units
OM = Perpendicular distance from O to
$2x-5y+18=0=\frac{2\times 3-5\times (-1)+18}{\sqrt{2^2+5^2}}=\frac{29}{\sqrt{29}}=\sqrt{29}$
Now, in right angle triangle AOB
$A{O}^2=O{M}^2+A{M}^2=29+3^2=38$
$\therefore AO=\sqrt{38}$ (radius)
Thus the equation of circle will be
$(x-3{)}^2+(y+1{)}^2=38\Rightarrow x^2+y^2-6x+2y-28=0$