If the line $l x + m y + n = 0$ touches the circle $x^{2} + y^{2} = a^{2}$ , then prove that $(l^{2} + m^{2}) a^{2} = n^{2}$ .

Open in App

Solution

If the line $l x + m y + n = 0$ touches the circle $x^{2} + y^{2} = a^{2}$ , then length of the perpendicular from its centre $O (0, 0)$ is equal to its radius a.

Therefore,
$| \frac{l \times 0 + m \times 0 + n}{\sqrt{l^{2} + m^{2}}} | = a$

$(l^{2} + m^{2}) a^{2} = n^{2}$ , which is the required condition.

Suggest Corrections

Similar questions

l x + m y + n = 0

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

\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}} = \frac{(a^{2} - b^{2})^{2}}{n^{2}}

l x + m y + n = 0

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

\frac{a^{2}}{l^{2}} - \frac{b^{2}}{m^{2}} = {(\frac{a^{2} + b^{2}}{n})}^{2}

l x + m y + n = 0

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

4 l^{2} + 3 m^{2} + n^{2} + 2 m n = 0

| g + f + c |

l x + m y = n

\frac{a^{2}}{l^{2}} + \frac{b^{2}}{m^{2}} = \frac{(a^{2} - b^{2})^{2}}{n^{2}}

l x + m y + n = 0

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{25} = 1

\frac{π}{2}

\frac{a^{2} l^{2} + b^{2} m^{2}}{n^{2}}

Join BYJU'S Learning Program