Question

Find the equation of the circle passing through the points :
(i) (5, 7), (8, 1) and (1, 3)
(ii) (1, 2), (3, -4), and (5, -6)
(iii) (5, -8), (-2, 9) and (2, 1)
(iv) (0, 0), (-2, 1) and (-3, 2)

Answer 1

We know that the general equation of circle is
$x^2+y^2+2gx+2fy+c=0\dots \left(i\right)$
We have,
$P(5,7),Q(8,1)$ and $R(1,3)$
Since P, Q and R lies on (i)
so,
$25+49+10g+14f+c=0\dots \left(ii\right)64+1+16g+2f+c=0\dots \left(iii\right)1+9+2g+6f+c=0\dots \left(iv\right)$
Solving (ii), (iii) and (iv), we get,
$g=-\frac{29}{6},f=\frac{19}{6},c=\frac{56}{3}$
Thus, the equation of circle is on putting g, f and c on (i)
$x^2+y^2-\frac{29}{3}x-\frac{19}{3}y+\frac{56}{3}=0\Rightarrow 3(x^2+y^2)-29x-19y+56=0$

We know that the general equation of circle is
$x^2+y^2+2gx+2fy+c=0\dots \left(1\right)$
We have, $P(1,2),Q(3,-4)$ and $R(5,-6)$
Since P, Q and R lies on (i)
so, $x^2+y^2+2gx+2fy+c=0\dots \left(i\right)1+4+2g+4f+c=0\dots \left(ii\right)9+16+6g-8f+c=0\dots \left(iii\right)25+36+lOg-12f+c=0\dots \left(iv\right)$
Solving (ii), (iii) and (iv), we get,
g = -11, = -2 and c =25\\
from (i)
The equation of circle is
$x^2+y^2-22x-4y+25=0$

(iii) We know that the general equation of circle is $x^2+y^2+2gx+2fy+c=0\dots \left(i\right)$
We have,
$P(5,-8),Q(-2,9)$ and $R=(2,1)$
P, Q and R lies on (i) so,
$25+64+lOg-16f+c=0\dots \left(ii\right)4+81-4g+18f+c=0\dots \left(iii\right)4+1+4g+2f+C=0\dots \left(iv\right)$
Solving (ii), (iii) and (iv), we get,
$g=58,f=24$ and $c=-285$
The equation of circle is
$x^2+y^2+116x+48y-285=0$
from (i)

(iv) We know that the general equation of circle is
$c^2+y^2+2gx+2fy+c=0\dots \left(1\right)$
We have,
$P(0,0),Q(-2,1)$ and $R(-3,2)$
Since P, Q and R. lies on (i)
so,
$0+00+0+c=0\dots \left(ii\right)4+1-4x+2y+c=0\dots \left(iii\right)9+4-6x+4y+c=0\dots \left(iv\right)$
Solving (ii), (iii) and (iv), we get,
$g=-\frac{3}{2},f=-\frac{11}{2},c=0$
from (i)
The equation of circle is
$x^2+y^2-3x-11y=0$