If the line `l x+m y-1=0` touches the circle `x^2+y^2=a^2` , then prove that `(l , m)` lies on a circle.

Explanation

Suppose if we let the line` lx+my−1=0` touches the circle `x^2+y^2=a^2`, then length of the perpendicular from its centre `C(0,0)` is equal to its radius `a`.
Then we have,
` | \frac{m+m\times 0-1}{\sqrt{l^{2}+m^{2}}} |=a`

`\Rightarrow l^{2}+m^{2}=\frac{1}{a^{2}}`

Hence, (l,m) lies on a circle.
Hence (l,m) lies on circle`\Rightarrow l^{2}+m^{2}=\frac{1}{a^{2}}`

then the point (l,m) satisfies the equation

`\Rightarrow x^{2}+y^{2}=\frac{1}{a^{2}}`

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