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

• A. \((x-2)^2+(y+1)^2 = 324\)
• B. \((x-2)^2+(y-1)^2 = 324\)
• C. \(5(x-2)^2+5(y+1)^2 = 324\)
• D. \(25(x-2)^2+25(y+1)^2 = 324\)

Hint:

The lines are parallel, so the diameter equals the distance between them and the centre is on the midline.

Answer 1

The Correct Option is D

Step 1: Approach
Let the centre be $(-2k,\,k)$ since it lies on $x+2y=0$, and impose equal distances to the two tangents.

Step 2: Distances
\[ \frac{|3(-2k)-4k+8|}{5}=\frac{|3(-2k)-4k-28|}{5} \]
\[ |8-10k|=|28+10k| \]
The case $8-10k=-(28+10k)$ gives $8=-28$, which is impossible. So $8-10k=28+10k$, giving $k=-1$ and the centre $(2,-1)$.

Step 3: Radius
\[ r=\frac{|6+4+8|}{5}=\frac{18}{5} \]

Step 4: Equation
\[ (x-2)^2+(y+1)^2=\frac{324}{25} \]
Multiplying by 25 gives option (D).

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). \[ \boxed{25(x-2)^2+25(y+1)^2=324} \]