Question:
Tangent to circle \(x^2+y^2-4x-8y-5=0\) equally inclined to axes is \(x+by+c=0\), \(b0\). Find \(2b+c\).
Tangent to circle \(x^2+y^2-4x-8y-5=0\) equally inclined to axes is \(x+by+c=0\), \(b0\). Find \(2b+c\).
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Equal inclination lines always have slope ±1.
Updated On: Jun 22, 2026
- \(4+5\sqrt2\)
- \(5\sqrt2\)
- \(-4-5\sqrt2\)
- \(-5\sqrt2\) \bigskip
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The Correct Option is C
Solution and Explanation
Concept:
Equally inclined line:
\[
|m|=1 \Rightarrow y=\pm x + c
\]
Step 1: Convert circle.
\[ (x-2)^2+(y-4)^2=25 \] Center \(C(2,4)\), radius \(5\)
Step 2: Use slope \(m=-1\).
Line: \[ x+y+c=0 \] Distance from center: \[ \frac{|6+c|}{\sqrt2}=5 \] \[ 6+c=\pm5\sqrt2 \] \[ c=-6\pm5\sqrt2 \]
Step 3: Compute.
\[ 2b+c=-2-6-5\sqrt2=-4-5\sqrt2 \] \[ \boxed{(C)} \]
Step 1: Convert circle.
\[ (x-2)^2+(y-4)^2=25 \] Center \(C(2,4)\), radius \(5\)
Step 2: Use slope \(m=-1\).
Line: \[ x+y+c=0 \] Distance from center: \[ \frac{|6+c|}{\sqrt2}=5 \] \[ 6+c=\pm5\sqrt2 \] \[ c=-6\pm5\sqrt2 \]
Step 3: Compute.
\[ 2b+c=-2-6-5\sqrt2=-4-5\sqrt2 \] \[ \boxed{(C)} \]
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