Q76 / 150 · MHT CET 2026 (11 April Shift 2) · Mathematics
If the lines $\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4}$ and $\dfrac{x-3}{1} = \dfrac{y-c}{2} = \dfrac{z}{1}$ intersect, then the radius of circle $x^2 + y^2 - 4x + 10y + c = 0$ is
- A$\dfrac{9}{\sqrt{2}}$
- B$\dfrac{9}{2}$
- Correct answerC$\dfrac{7}{\sqrt{2}}$
- D$\dfrac{7}{2}$
Solution
The condition for the two lines to intersect is that they must be coplanar.
$\begin{vmatrix} 3-1 & c-(-1) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$
$\begin{vmatrix} 2 & c+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(3 - 8) - (c+1)(2 - 4) - 1(4 - 3) = 0$
$-10 + 2(c+1) - 1 = 0$
$2c - 9 = 0 \Rightarrow c = \dfrac{9}{2}$
The equation of the circle is $x^2 + y^2 - 4x + 10y + \dfrac{9}{2} = 0$.
The radius of the circle is given by $r = \sqrt{g^2 + f^2 - c}$.
$r = \sqrt{(-2)^2 + (5)^2 - \dfrac{9}{2}}$
$r = \sqrt{4 + 25 - \dfrac{9}{2}}$
$r = \sqrt{29 - \dfrac{9}{2}} = \sqrt{\dfrac{49}{2}} = \dfrac{7}{\sqrt{2}}$
Answer: $\dfrac{7}{\sqrt{2}}$
$\begin{vmatrix} 3-1 & c-(-1) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$
$\begin{vmatrix} 2 & c+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(3 - 8) - (c+1)(2 - 4) - 1(4 - 3) = 0$
$-10 + 2(c+1) - 1 = 0$
$2c - 9 = 0 \Rightarrow c = \dfrac{9}{2}$
The equation of the circle is $x^2 + y^2 - 4x + 10y + \dfrac{9}{2} = 0$.
The radius of the circle is given by $r = \sqrt{g^2 + f^2 - c}$.
$r = \sqrt{(-2)^2 + (5)^2 - \dfrac{9}{2}}$
$r = \sqrt{4 + 25 - \dfrac{9}{2}}$
$r = \sqrt{29 - \dfrac{9}{2}} = \sqrt{\dfrac{49}{2}} = \dfrac{7}{\sqrt{2}}$
Answer: $\dfrac{7}{\sqrt{2}}$
Hence, the correct answer is option: 'C'.