If $ax + by = c$ is tangent to the circle $x^2 + y^2 = 16$

Question

Problem: If $ax + by = c$ is tangent to the circle $x^2 + y^2 = 16$ then which of the following is correct option

(A) $16 ( a^2 + b^2) = c ^2 $

(B) $16 ( a^2 - b^2) = c ^2 $

(C) $16 ( a^2 +b^2) = - c^2 $

(D) $16 ( a^2 - b^2) = - c^2$

Solution:

As the line is a tangent there will be a single solution for the set of equation.

Substitute either x or y from the linear equation in the quadratic equation and use condition both the roots are equal i.e $D=0$

We will get answer as option (A)

Is above method is correct

Roots are equal i.e $D=0$ .But it will form equation of degree $4$ either in $a$ or $b$ .How will we get option (A) — rst, Jun 18 '14 at 09:08
A simpler way of doing this is to set the distance of this line from the center (0,0) equal to the radius. — Anurag A, Jun 18 '14 at 09:10
Your method is correct. And when you get the fourth degree equation, it factors as $(16(a^2+b^2)-c^2)\cdot a^2$, or similar. — , Jul 11 '14 at 10:08

score 2 · Answer 1 · edited Jun 20 '14 at 11:33

2

$$x^2+y^2=16 \implies xdx+ydy=0 \implies \frac{dy}{dx}=-\frac{x}{y}$$ So the slope of the tangent to the circle at point $(x_0,y_0)$ is just $-\frac{x_0}{y_0}$.

A line tangent to the circle at point $(x_0,y_0)$ is given by:

$$y-y_0=-\frac{x_0}{y_0}(x-x_0) \implies x_0 x+y_0 y=x_0^2+y_0^2$$

Thus $a=x_0,b=x_0,c=x_0^2+y_0^2=16$

So option (A) is the only right answer.

edited Jun 20 '14 at 11:33

Debashish

1,714

answered Jun 18 '14 at 09:35

mike

5,604

score 0 · Answer 2 · answered Jul 11 '14 at 09:34

0

By symmetry ($a$ and $b$ are interchangeable), $(B)$ and $(D)$ must be excluded. And $(C)$ is an incompatible expression.

So $(A)$, without further analysis.

answered Jul 11 '14 at 09:34

If $ax + by = c$ is tangent to the circle $x^2 + y^2 = 16$

2 Answers2