The line lx - my - n -0 will be a tangent to the circle x 2- y 2- a 2 iff A- n212-m2-a2B- a 212- m 2- n 2C- n 1- m - aD- a 1- m - n

The correct option is B a^2(l^2 + m^2) = n^2Line y = mx + c is tangent, if c = ± a√(1+m^2). Now lx + my + n = 0 or y = 1/m x n/m is tangent, if n/m = ± ...

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Standard XII

Mathematics

Equation of Tangent in Slope Form

The line lx +...

Question

The line lx + my + n = 0 will be a tangent to the circle $x^{2} + y^{2} = a^{2}$ iff

n^{2} (l^{2} + m^{2}) = a^{2}

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a^{2} (l^{2} + m^{2}) = n^{2}

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n(l + m) = a

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a(l + m) = n

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Solution

The correct option is B $a^{2} (l^{2} + m^{2}) = n^{2}$
Line y = mx + c is tangent, if $c = \pm a \sqrt{1 + m^{2}} .$
Now lx + my + n = 0 or $y = - \frac{1}{m} x - \frac{n}{m}$ is tangent, if $- \frac{n}{m} = \pm a \sqrt{1 + {(\frac{1}{m})}^{2}} o r n^{2} = a^{2} (m^{2} + l^{2}) .$

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The line lx + my + n = 0 will be a tangent to the circle x2+y2=a2 iff

The correct option is B a2(l2+m2)=n2Line y = mx + c is tangent, if c=± a√1+m2. Now lx + my + n = 0 or y=−1mx−nm is tangent, if −nm=±a√1+(1m)2 or n2=a2(m2+l2).

The line lx + my + n = 0 will be a tangent to the circle $x^{2} + y^{2} = a^{2}$ iff

The correct option is B $a^{2} (l^{2} + m^{2}) = n^{2}$
Line y = mx + c is tangent, if $c = \pm a \sqrt{1 + m^{2}} .$
Now lx + my + n = 0 or $y = - \frac{1}{m} x - \frac{n}{m}$ is tangent, if $- \frac{n}{m} = \pm a \sqrt{1 + {(\frac{1}{m})}^{2}} o r n^{2} = a^{2} (m^{2} + l^{2}) .$