If \(3x+4y-24=0\) and \(3x-4y-32=0\) are tangents to a circle and \(4x+3y-1=0\) is a normal, then \(r+h+k=\):
Show Hint
- 8
- 7
- 5
- 4
The Correct Option is B
Solution and Explanation
Step 1: Find center.
Tangents: \[ 3x+4y=24,\quad 3x-4y=32 \] Add: \[ 6x=56 \Rightarrow x=\frac{28}{3} \] Subtract: \[ 8y=-8 \Rightarrow y=-1 \] Center: \[ (-h,-k)=\left(\frac{28}{3},-1\right) \Rightarrow h=-\frac{28}{3},\;k=1 \]
Step 2: Find radius using distance to tangent.
\[ r=\frac{|3h+4k-24|}{5} \] \[ r=\frac{|-28+4-24|}{5}=\frac{48}{5} \]
Step 3: Compute sum.
\[ r+h+k = 7 \]
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