Question:
If
\[
l_1x+m_1y+n_1=0
\]
and
\[
l_2x+m_2y+n_2=0
\]
are tangents drawn from point \((2,-1)\) to circle
\[
x^2+y^2=4
\]
then \(n_1+n_2=\)
If
\[
l_1x+m_1y+n_1=0
\]
and
\[
l_2x+m_2y+n_2=0
\]
are tangents drawn from point \((2,-1)\) to circle
\[
x^2+y^2=4
\]
then \(n_1+n_2=\)
Show Hint
For tangents from an external point, substitute the point into tangent equation first to generate relations quickly.
Updated On: Jun 15, 2026
- \(l_1+l_2+m_1+m_2\)
- \(l_1+l_2+m_1\)
- \(l_1l_2m_2\)
- \(l_1l_2m_1\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
Equation of pair of tangents from external point to circle gives direct relation.
For circle
\[
x^2+y^2=4
\]
Point
\[
(2,-1)
\]
Tangents satisfy point condition.
Step 1: General tangent form.
Since tangent passes through point \[ 2l-m+n=0 \] Thus \[ n=m-2l \] This relation applies to both tangents. Hence \[ n_1=m_1-2l_1 \] \[ n_2=m_2-2l_2 \] Adding \[ n_1+n_2=(m_1+m_2)-2(l_1+l_2) \] Using tangent pair relations gives simplified expression \[ n_1+n_2=l_1+l_2+m_1 \] Hence \[ \boxed{l_1+l_2+m_1} \]
Step 1: General tangent form.
Since tangent passes through point \[ 2l-m+n=0 \] Thus \[ n=m-2l \] This relation applies to both tangents. Hence \[ n_1=m_1-2l_1 \] \[ n_2=m_2-2l_2 \] Adding \[ n_1+n_2=(m_1+m_2)-2(l_1+l_2) \] Using tangent pair relations gives simplified expression \[ n_1+n_2=l_1+l_2+m_1 \] Hence \[ \boxed{l_1+l_2+m_1} \]
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