Question

Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y - 4x + 3 = 0.

Answer 1

Let the general equation of the circle is
$x^2+y^2+2gx+2fy+c=0\dots \left(i\right)$
Since, this circle passes through the points (2, 3) and (4, 5).
$\therefore 4+9+4g+6f+c=0\Rightarrow 4g+6f+c=-13\dots \left(ii\right)$
and $16+25+8g+10f+c=0\Rightarrow 8g+10f+c=-41\dots \left(iii\right)$
Since, the centre of the circle (-g, -f) lies on the straight line y 4x + 3 = 0
i.e., $+4g-f+3=0\dots \left(iv\right)$
From Eq. (iv), 4g = f- 3
On putting 4g =f- 3 in Eq. (ii), we get
$f-3+6f+c=-13\Rightarrow 7f+c=10$
From Eqs. (ii) and (iii),

$8g+17f+2c=-268g+10f+2c=-41\underset{–}{---+}\underset{–}{2f+c=15}\dots \left(vi\right)$
From Eqs. (ii) and (vi)
$7f+c=-102f+c=15\underset{–}{---}\underset{–}{5f=-25}\therefore f=-5$
Now, $c=10+15=25$
From Eq. (iv), $4g+5+3=0$
$\Rightarrow g=-2$
From Eq. (i), equation of the circle is $x^2+y^2-4x-10y+25=0$