Question:
The equation of a circle whose center is at \((3,-4)\) and the line \(12x-5y-17=0\) cuts the circle at \(A\) and \(B\), where \(l(AB)=6\) is ...
The equation of a circle whose center is at \((3,-4)\) and the line \(12x-5y-17=0\) cuts the circle at \(A\) and \(B\), where \(l(AB)=6\) is ...
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Find the distance from the centre to the line, then use r squared = d squared + (half chord) squared.
Updated On: Nov 12, 2026
- \(x^{2}+y^{2}-6x+8y+7=0\)
- \(x^{2}+y^{2}-6x+8y-7=0\)
- \(x^{2}+y^{2}+6x+8y+7=0\)
- \(x^{2}+y^{2}-6x-8y-7=0\)
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The Correct Option is A
Solution and Explanation
Step 1: Distance from centre to line
\(d=\dfrac{|12(3)-5(-4)-17|}{\sqrt{144+25}}=\dfrac{39}{13}=3\)
Step 2: Use the chord property
Half chord \(=3\), so \(r^2=d^2+3^2=9+9=18\).
Step 3: Equation
\((x-3)^2+(y+4)^2=18\)
Step 4: Expand
\(x^2+y^2-6x+8y+25-18=0\Rightarrow x^2+y^2-6x+8y+7=0\)
Final Answer:
The circle is x squared plus y squared minus 6x plus 8y plus 7 equals 0. \[ \boxed{\text{(A) }x^{2}+y^{2}-6x+8y+7=0} \]
\(d=\dfrac{|12(3)-5(-4)-17|}{\sqrt{144+25}}=\dfrac{39}{13}=3\)
Step 2: Use the chord property
Half chord \(=3\), so \(r^2=d^2+3^2=9+9=18\).
Step 3: Equation
\((x-3)^2+(y+4)^2=18\)
Step 4: Expand
\(x^2+y^2-6x+8y+25-18=0\Rightarrow x^2+y^2-6x+8y+7=0\)
Final Answer:
The circle is x squared plus y squared minus 6x plus 8y plus 7 equals 0. \[ \boxed{\text{(A) }x^{2}+y^{2}-6x+8y+7=0} \]
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