Question

The line y=mx + c is tangent to the circle x² + y² = 49. The correct condition on c is:

• A. c = ±9√1 + m²
• B. c = ±11√1 + m²
• C. c = ±13√1 + m²
• D. c = ±7√1 + m²

Answer 1

Answer: (D)

c = ±7√1 + m²

Answer 2

The problem asks for the condition on 'c' for the line $y = mx + c$ to be tangent to the circle $x^2 + y^2 = 49$.

1. Identify the circle's properties:

The given equation of the circle is $x^2 + y^2 = 49$. This is in the standard form $(x-h)^2 + (y-k)^2 = r^2$, where $(h,k)$ is the center and $r$ is the radius. Comparing, we find: Center $(h,k) = (0,0)$ Radius $r^2 = 49 \Rightarrow r = 7$

2. Rewrite the line equation:

The equation of the line is $y = mx + c$. To use the distance formula, we rewrite it in the general form $Ax + By + C = 0$: $mx - y + c = 0$ Here, $A = m$, $B = -1$, $C = c$.

3. Apply the tangency condition:

A line is tangent to a circle if and only if the perpendicular distance from the center of the circle to the line is equal to the radius of the circle. The formula for the perpendicular distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

For our case, $(x_1, y_1) = (0,0)$ (the center of the circle), $A = m$, $B = -1$, $C = c$. Substituting these values: $d = \frac{|m(0) - 1(0) + c|}{\sqrt{m^2 + (-1)^2}}$ $d = \frac{|c|}{\sqrt{m^2 + 1}}$

For tangency, this distance $d$ must be equal to the radius $r=7$: $\frac{|c|}{\sqrt{m^2 + 1}} = 7$

4. Solve for c:

$|c| = 7\sqrt{m^2 + 1}$ This implies $c$ can be positive or negative: $c = \pm 7\sqrt{1 + m^2}$

Comparing this with the given options, the correct condition is $c = \pm 7\sqrt{1 + m^2}$.