The equation of the circle passing through the point $(1,2)$ and through the points of intersection of…
The equation of the circle passing through the point $(1,2)$ and through the points of intersection of $x^2+y^2-4 x-6 y-21=0$ and $3 x+4 y+5=0$ is given by
$x^2+y^2+2 x+2 y+11=0$
$x^2+y^2-2 x+2 y-7=0$
$x^2+y^2+2 x-2 y-3=0$
$x^2+y^2+2 x+2 y-11=0$
Solution
Point $(1,2)$ lies on the circle $x^2+y^2+2 x+2 y$ $-11=0$, because coordinates of point $(1,2)$ satisfy the equation
$
x^2+y^2+2 x+2 y-11=0
$
Now, $x^2+y^2-4 x-6 y-21=0 \quad$...(i)
$
\begin{aligned}
& x^2+y^2+2 x+2 y-11=0 \\
& 3 x+4 y+5=0
\end{aligned}
$
From (i) and (iii),
$
\begin{array}{r}
x^2+\left(-\frac{3 x+5}{4}\right)^2-4 x-6\left(-\frac{3 x+5}{4}\right)-21=0 \\
\Rightarrow \quad 16 x^2+9 x^2+30 x+25-64 x \\
+72 x+120-336=0
\end{array}
$
$
\begin{aligned}
& \Rightarrow \quad 25 x^2+38 x-191=0 \\
& \text { From (ii) and (iii), } \\
& x^2+\left(-\frac{3 x+5}{4}\right)^2+2 x+2\left(-\frac{3 x+5}{4}\right)-11=0 \\
& \Rightarrow \quad 16 x^2+9 x^2+30 x+25+32 x
\end{aligned}
$
Thus we get the same equation from (ii) and (iii) as we get from equation (i) and (iii). Hence the point of intersections of (ii) and (iii) will be same as the point of intersections of(i) and (iii). Therefore the circle (ii) passing through the point of intersection of circle(i) and point $(1,2)$ also as shown in the figure.
Hence equation(ii) i.e. $x^2+y^2+2 x+2 y-11=0$ is the equation of required circle.
