Question

If the equations of two diameters of a circle arc 2x + y = 6 and 3x + 2y = 4 and the radius is 10, find the equation of the circle.

Answer 1

The equation of two diameter of the circle
$(x-a{)}^2+(y-b{)}^2=r^2\dots \left(A\right)is2x+y=6\dots \left(i\right)3x+2y=4\dots \left(ii\right)$
By solving (i) and (ii) we will get
$x=8,y=-10$
$4x+2y=123x+2y=4\underset{–}{---}\underset{–}{x=+8}$
Also $2\left(8\right)+y=616+y=6y=6-16y=-10$
The point of intersection of (l) and (2) is C (8, -10), which is the centre of circle. Also, radius = 10
$\therefore$ From (A)
$\Rightarrow (x-8{)}^2+(y+10{)}^2={10}^2\Rightarrow x^2+y^2-16x+20y+64=0$