Question

Find the equation of a circle
which touches both the axes at a distance of 6 units from the origin.

Answer 1

Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be

\[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]

(i) Let the required equation of the circle be

\[\left( x - h \right)^2 + \left( y - k \right)^2 = a^2\]

It is given that the circle passes through the points (6, 0) and (0, 6).

∴\[\left( 6 - h \right)^2 + \left( 0 - k \right)^2 = 6^2\]

And,

\[\left( 0 - h \right)^2 + \left( 6 - k \right)^2 = 6^2\]

\[\Rightarrow \left( 6 - h \right)^2 + \left( - k \right)^2 = 36\]
\[ \Rightarrow 36 + h^2 - 12h + k^2 = 36\]
\[ \Rightarrow h^2 + k^2 = 12h . . . (1)\]

Also,

\[h^2 + 36 + k^2 - 12k = 36\]

\[\Rightarrow h^2 + k^2 = 12k\] ...(2)

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

\[12k = 12h \Rightarrow h = k\]

∴ From equation (2), we have:

\[k^2 + k^2 = 12k\]
\[ \Rightarrow k^2 - 6k = 0\]
\[ \Rightarrow k\left( k - 6 \right) = 0\]
\[ \Rightarrow k = 6 \left( \because k > 0 \right)\]

Consequently, we get:
h = 6
Hence, the required equation of the circle is

\[\left( x - 6 \right)^2 + \left( y - 6 \right)^2 = 36\]

\[x^2 + y^2 - 12x - 12y + 36 = 0\]