Question

If a straight line y = mx + c touches a fixed circle such that $21m^2 - c^2 + 12m + 6c - 4mc + 16 = 0$, then radius of the circle is equal to-

• A. 2
• B. 3
• C. 5
• D. 7

Answer 1

Answer: (C)

5

Answer 2

The condition for a line $y=mx+c$ to be tangent to a circle with center $(h, k)$ and radius $r$ is that the perpendicular distance from the center to the line is equal to the radius. The distance is given by $d = \frac{|mh - k + c|}{\sqrt{m^2 + 1}}$. For tangency, $d=r$, so $r^2(m^2+1) = (mh - k + c)^2$. Expanding and rearranging this equation to be a quadratic in $c$: $c^2 + (2mh - 2k)c + m^2(h^2 - r^2) - 2mhk + k^2 - r^2 = 0$.

The given condition is $21m^2 - c^2 + 12m + 6c - 4mc + 16 = 0$. Rearranging this as a quadratic in $c$: $-c^2 + (6 - 4m)c + (21m^2 + 12m + 16) = 0$ Multiplying by $-1$: $c^2 + (4m - 6)c - (21m^2 + 12m + 16) = 0$.

Comparing the coefficients of the two quadratic equations in $c$:

1. Coefficient of $c$: $2mh - 2k = 4m - 6$ Equating coefficients of $m$: $2h = 4 \implies h = 2$. Equating constant terms: $-2k = -6 \implies k = 3$. The center of the circle is $(2, 3)$.

2. Constant term (independent of $c$): $m^2(h^2 - r^2) - 2mhk + k^2 - r^2 = -(21m^2 + 12m + 16)$ Substitute $h=2$ and $k=3$: $m^2(2^2 - r^2) - 2(2)(3)m + 3^2 - r^2 = -21m^2 - 12m - 16$ $m^2(4 - r^2) - 12m + 9 - r^2 = -21m^2 - 12m - 16$

   Equating coefficients of $m^2$: $4 - r^2 = -21$ $r^2 = 4 + 21$ $r^2 = 25$ $r = 5$ (since radius must be positive).

   The coefficients of $m$ and the constant terms are consistent with these values. Thus, the radius of the circle is 5.