Question

Find the equation of the circle passing through the points A (1 , 2) , B (3 , 4 ) and tangent to the straight lines 3x + y - 3 = 0

Answer 1

Let C (h,K) be the centre and A,B be the given points, then
$C{A}^2=C{B}^2=r^2=p^2$
Where p is perpendicular from centre to the tangent $3x+y-3=0$
$C{A}^2=C{B}^2\Rightarrow h+k=5$
$C{A}^2=p^2\Rightarrow {(h-1)}^2+{(k-2)}^2$
$=\left(\frac{3h+k-3}{\sqrt{10}}\right)$
Now eliminate k and you wil find
$2h^2-11h+12=10$
Hence the circles are $(x-h{)}^2+(y-k{)}^2=r^2$