Question

Find the equation of the circle having (1, −2) as its centre and passing through the intersection of the lines 3x + y = 14 and 2x + 5y = 18. [NCERT EXEMPLAR]

Answer 1

Solving 3x + y = 14 and 2x + 5y = 18 we get
x = 4 and y = 2
The radius is equal to the distance between (1, −2) and (4, 2)
$$\begin{gathered} r=\sqrt{{\left(4-1\right)}^2+{\left(2+2\right)}^2} \\ =\sqrt{9+16} \\ =5 \end{gathered}$$
Now, the equation of the circle is given by
$$\begin{gathered} {\left(x-h\right)}^2+{\left(y-k\right)}^2=r^2 \\ \Rightarrow {\left(x-1\right)}^2+{\left(y+2\right)}^2=25 \\ \Rightarrow x^2+y^2-2x+4y-20=0 \end{gathered}$$