Question

Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).

Answer 1

We know that the equation of circle whose ccntrd in (a, b) and radius r is
$(x-a{)}^2+(y-b{)}^2=r^2\dots \left(1\right)$
We have centre - (1, 2)
$\therefore (x-1{)}^2+(y-2{)}^2=r^2\dots \left(2\right)$
Also, circle passes through (4, 6)
$\therefore (4-1{)}^2+(6-2{)}^2=r^2\Rightarrow 9+16=r^2\Rightarrow r=5$
Thus, equation of required circle in
$(x-1{)}^2+(y-2{)}^2=5^2{x}^2+y^2-2x-4y-20=0$