Question

Find the equation of the circle passing through the point (4,1) and (6,5) and whose centre is on the line 4x+y = 16

Answer 1

The equation of te ircle is
$(x-h{)}^2+(y-k{)}^2=r^2$

Since th circle passes through point (4,1)
$\therefore (4-h{)}^2+(1-k{)}^2=r^2$
$\Rightarrow 16+h^2-8h+1+k^2-2k=r^2\Rightarrow h^2+k^2-8h-2k+17=r^2$
Also the circle passes through point (6,5)
$\therefore (6-h{)}^2+(5-k{)}^2=r^2\Rightarrow 36+h^2-12h+25+k^2-10k=r^2\Rightarrow h^2+k^2-12hy-10k+61=r^2$
From (ii) and (ii) we have
$h^2+k^2-8h-2k+17=h^2+k^2-12h-10k+61\Rightarrow 4h+8k=44\Rightarrow h+2k=11$
Since the centre (h, k) of the circle lies on the
line 4x+y = 16
$\therefore 4h+k=16$
Solving (iv) and (v) we have h = 3 and k = 4 putting value of h and k in (ii) we have
$(3{)}^2+(4{)}^2-8\times 3-2\times 4+17=r^2\therefore r^2=10$
thus equation of required circle is
$(x-3{)}^2+(y-4{)}^2=10\Rightarrow x^2+9-6x+y^{2}16-8y=10\Rightarrow x^2+y^2-6x-8y+15=0$