If the lines frac{x - 1}{2} = frac{y + 1}{3} | vedclass

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(C) Step $1$: Find the intersection condition for the lines. Let the points on the lines be $(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$ and $(\mu + 3

Vedclass · Accepted Answer

$\frac{7}{\sqrt{2}}$

Vedclass · Answer

(C) Step $1$: Find the intersection condition for the lines. Let the points on the lines be $(2\lambda + 1, 3\lambda - 1, 4\lambda + 1)$ and $(\mu + 3, 2\mu + c, \mu)$.Step $2$: Equating coordinates: $2\lambda + 1 = \mu + 3 \implies 2\lambda - \mu = 2$ $(i)$, $3\lambda - 1 = 2\mu + c \implies 3\lambda - 2\mu = c + 1$ (ii), $4\lambda + 1 = \mu \implies 4\lambda - \mu = -1$ (iii).Step $3$: Solving $(i)$ and (iii): Subtracting $(i)$ from (iii) gives $2\lambda = -3 \implies \lambda = -1.5$. Then $\mu = 4(-1.5) + 1 = -5$.Step $4$: Substitute $\lambda, \mu$ into (ii): $3(-1.5) - 2(-5) = c + 1 \implies -4.5 + 10 = c + 1 \implies 5.5 = c + 1 \implies c = 4.5 = \frac{9}{2}$.Step $5$: The circle equation is $x^2 + y^2 - 4x + 10y + 4.5 = 0$. Comparing with $x^2 + y^2 + 2gx + 2fy + c' = 0$, we have $g = -2, f = 5, c' = 4.5$.Step $6$: Radius $r = \sqrt{g^2 + f^2 - c'} = \sqrt{(-2)^2 + 5^2 - 4.5} = \sqrt{4 + 25 - 4.5} = \sqrt{24.5} = \sqrt{\frac{49}{2}} = \frac{7}{\sqrt{2}}$.

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 the circle $x^2 + y^2 - 4x + 10y + c = 0$ is

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