Question

Find the equation to the circle which passes through the points (1, 1) (2, 2) and whose radius is 1. Show that there are two such circles.

Answer 1

Let the equation of the required circle be

\[x^2 + y^2 + 2gx + 2fy + c = 0\]

It passes through (1, 1) and (2, 2).

∴ \[2g + 2f + c = - 2\]...(1)

And,

\[4g + 4f + c = - 8\] ...(2)

From (1) and (2), we have:

\[- 2g - 2f = 6 \Rightarrow g + f = - 3\] ...(3)

∴ From (2) and (3), we have:

\[c = 4\]

\[Also, \sqrt{g^2 + f^2 - c} = 1\]
\[ \Rightarrow g^2 + f^2 = 1 + c = 5\]
\[ \Rightarrow \left( g + f \right)^2 - 2gf = 5\]
\[ \Rightarrow gf = 2\]

Using (3), we get:

\[g = - 2, - 1\]

Correspondingly, we have:

\[f = - 1, - 2\]

Therefore, the required equations of the circles are

\[x^2 + y^2 - 4x - 2y + 4 = 0\]

\[x^2 + y^2 - 2x - 4y + 4 = 0\]

Hence, there are two such circles.