Question

if the line lx+my+n=0 touches the circle x²+y²=r², then prove that (l²+m²)r²=n²

Answer 1

The answer is given below :

The circle is : x² + y² = r² .....(ii)

and

the line is : lx + my + n = 0 .....(ii)

The centre of the circle (i) is at (0, 0) and radius is r units.

Since the line (ii) touches the circle (i), the radius of the circle and the perpendicular distance from the centre (0, 0) of the circle to the line (ii) are equal.

Now, the perpendicular distance of the line lx + my + n = 0 from the point (0, 0) is

$= \frac{l \times 0 + m \times 0 + n}{ \sqrt{ {l}^{2} + {m}^{2} } } \\ \\ = \frac{n}{ \sqrt{ {l}^{2} + {m}^{2} } }$

By the given condition,

$r = \frac{n}{ \sqrt{ {l}^{2} + {m}^{2} } } \\ \\ = > \: r \sqrt{ {l}^{2} + {m}^{2} } = n \\ \\ squaring \: \: we \: \: get \\ \\ {r}^{2} ( {l}^{2} + {m}^{2} ) = {n}^{2} \\ \\ hence \: \: proved$

Thank you for your question.

Answer 2

Answer:

Answer is l/(1/3) = m/(-1/2) = -n/(7). Hence 3l=-2m=-n/7