Question:
The equation of a circle whose center lies on \(x+2y = 0\) and touching the lines \(3x-4y+8 = 0\) and \(3x-4y-28 = 0\) is
The equation of a circle whose center lies on \(x+2y = 0\) and touching the lines \(3x-4y+8 = 0\) and \(3x-4y-28 = 0\) is
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The lines are parallel, so the diameter equals the distance between them and the centre is on the midline.
Updated On: Oct 1, 2026
- \((x-2)^2+(y+1)^2 = 324\)
- \((x-2)^2+(y-1)^2 = 324\)
- \(5(x-2)^2+5(y+1)^2 = 324\)
- \(25(x-2)^2+25(y+1)^2 = 324\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
The two lines \(3x-4y+8=0\) and \(3x-4y-28=0\) are parallel. A circle touching both has its diameter equal to the distance between them and its centre on the line midway between them.
Step 2: Radius
\[ d=\frac{|8-(-28)|}{\sqrt{3^2+4^2}}=\frac{36}{5},\qquad r=\frac{18}{5},\quad r^2=\frac{324}{25} \]
Step 3: Locus of the centre
The centre is equidistant from both lines, so it lies on the mid line:
\[ 3x-4y+\frac{8-28}{2}=0\Rightarrow 3x-4y-10=0 \]
Step 4: Solve with the given line
\(x+2y=0\Rightarrow x=-2y\). Substitute:
\[ -6y-4y=10\Rightarrow y=-1,\quad x=2 \]
Step 5: Equation
\[ (x-2)^2+(y+1)^2=\frac{324}{25}\ \Rightarrow\ 25(x-2)^2+25(y+1)^2=324 \]
This is option (D). Options (A) and (B) have the radius squared as 324, which is 25 times too large.
Final Answer:
The centre is (2, -1) and the radius is 18/5, giving \(25(x-2)^2+25(y+1)^2=324\), option (D). \[ \boxed{25(x-2)^2+25(y+1)^2=324} \]
The two lines \(3x-4y+8=0\) and \(3x-4y-28=0\) are parallel. A circle touching both has its diameter equal to the distance between them and its centre on the line midway between them.
Step 2: Radius
\[ d=\frac{|8-(-28)|}{\sqrt{3^2+4^2}}=\frac{36}{5},\qquad r=\frac{18}{5},\quad r^2=\frac{324}{25} \]
Step 3: Locus of the centre
The centre is equidistant from both lines, so it lies on the mid line:
\[ 3x-4y+\frac{8-28}{2}=0\Rightarrow 3x-4y-10=0 \]
Step 4: Solve with the given line
\(x+2y=0\Rightarrow x=-2y\). Substitute:
\[ -6y-4y=10\Rightarrow y=-1,\quad x=2 \]
Step 5: Equation
\[ (x-2)^2+(y+1)^2=\frac{324}{25}\ \Rightarrow\ 25(x-2)^2+25(y+1)^2=324 \]
This is option (D). Options (A) and (B) have the radius squared as 324, which is 25 times too large.
Final Answer:
The centre is (2, -1) and the radius is 18/5, giving \(25(x-2)^2+25(y+1)^2=324\), option (D). \[ \boxed{25(x-2)^2+25(y+1)^2=324} \]
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