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: Use the intersection condition:
Two lines through points $P_1(1,-1,1)$, $P_2(3,c,0)$ with directions $\vec d_1=(2,3,4)$, $\vec d_2=(1,2,1)$ intersect when $(\vec P_2-\vec P_1)\cdot(\vec d_1\times\vec d_2)=0$.

Step 2: Compute:
$\vec d_1\times\vec d_2=(3-8,\ 4-2,\ 4-3)=(-5,2,1)$. $\vec P_2-\vec P_1=(2,c+1,-1)$. Dot product: $-10+2(c+1)-1=0$, so $2c=9$ and $c=\dfrac92$.

Step 3: Radius:
$r^2=4+25-\dfrac92=\dfrac{49}{2}$, so $r=\dfrac{7}{\sqrt2}$. Option C.

Final Answer:
Intersection gives c = 9/2, so r squared is 49/2. \[ \boxed{\text{(C) }\dfrac{7}{\sqrt2}} \]