Question

If the lines \(\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}\) and \(\frac{x-3}{1} = \frac{y-c}{2} = \frac{z}{1}\) intersect, then the radius of circle \(x^2+y^2-4x+10y+c = 0\) is

• A. \(\frac{9}{\sqrt{2}}\)
• B. \(\frac{9}{2}\)
• C. \(\frac{7}{\sqrt{2}}\)
• D. \(\frac{7}{2}\)

Hint:

Make the two lines meet to find c, then find the radius from c.

Answer 1

The Correct Option is C

Step 1: Parametrise the lines:
Line 1: \((1+2s,\ -1+3s,\ 1+4s)\). Line 2: \((3+t,\ c+2t,\ t)\).

Step 2: Equate x and z:
\(1+2s=3+t\) and \(1+4s=t\). Substitute \(t=1+4s\): \(1+2s=4+4s\), so \(s=-\dfrac32\) and \(t=1-6=-5\).

Step 3: Equate y to find c:
\(-1+3s=-1-\dfrac92=-\dfrac{11}{2}\) and \(c+2t=c-10\). So \(c-10=-\dfrac{11}{2}\), giving \(c=\dfrac92\).

Step 4: Find the radius:
For \(x^2+y^2-4x+10y+c=0\): \(g=-2\), \(f=5\). \(r=\sqrt{g^2+f^2-c}=\sqrt{4+25-\tfrac92}=\sqrt{\dfrac{49}{2}}=\dfrac{7}{\sqrt2}\). Option C.

Step 5: Why the other options are wrong.
\(\dfrac{9}{\sqrt2}\), \(\dfrac92\) and \(\dfrac72\) would need different values of \(c\). They come from a wrong sign of \(c\) (using \(-\tfrac92\)) or from forgetting the square root in \(r\).

Final Answer:
The radius is 7 / sqrt 2. \[ \boxed{\text{(C) }\dfrac{7}{\sqrt2}} \]