The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if:

Question

$1$. $c=a\sqrt {1+m^2}$

$2$. $c=\pm a\sqrt {1+m^2}$

$3$. $c^2=\pm a\sqrt {1+m^2}$

$4$. $\textrm {None}$

My Attempt:

The tangent to circle $x^2+y^2=a^2$ at point $(x_1,y_1)$ is given by : $$xx_1+yy_1=a^2$$

Now, what should I do next?

also $y_1 = mx_1+c$ and now you have two equations to solve. — dezdichado, Apr 27 '17 at 01:47
@labbhattacharjee, With centre $(0,0)$ and radius$=r$, I got: $$a=\dfrac {c}{\sqrt {m^2+1}}$$. So whats next? — pi-π, Apr 27 '17 at 12:40
@Maxwell, It should be $$|a|=\dfrac c{\sqrt{m^2+1}}\implies c=?$$ — lab bhattacharjee, Apr 27 '17 at 12:48

score 1 · Accepted Answer · answered Apr 27 '17 at 01:53

1

You have $x^2 + (mx+c)^2 = a^2 \iff x^2 + m^2x^2 + 2mxc + c^2 = a^2$ which simplifies to $$(1+m^2)x^2 + 2mcx + (c^2-a^2) = 0$$

But as $y$ must be tangent, there must be a single solution to the above quadratic. Which means its discriminant must be $0$, i.e: $$4m^2c^2-4(1+m^2)(c^2-a^2) = 0$$

answered Apr 27 '17 at 01:53

Zain Patel

16,802

what does it mean " there must be a single solution to the above quadratic."? – pi-π Apr 29 '17 at 12:48
and WHY is so, for tangency? – pi-π Apr 29 '17 at 12:48
He meant the roots coincide...(for tangency and single solution) – Narasimham Oct 27 '17 at 20:12

score 0 · Answer 2 · answered Apr 27 '17 at 14:20

A possible approach:

1) Calculate distance from (0,0) to line $y = mx + c$.

The perpendicular line passing through the origin is

$y = - (1/m)x$.

Point of intersection: $- (1/m)x = mx + c$.

$- x = m^2x + mc$

$x ( m^2 + 1) + mc = 0$;

$x = (-mc)/(m^2 +1)$;

Corresponding $y$:

$y = - (1/m) (-mc) /(m^2 + 1)$;

$y = c/(m^2 +1)$.

Distance (squared) to the origin:

$x^2 + y^2 = [(mc)^2 + c^2]/(m^2 +1)^2$ =

= $c^2(m^2 +1)/(m^2 + 1)^2 $=

$c^2/(m^2 + 1)$.

2) For this point to be a point of tangency:

Distance (squared) = radius (squared) =

$a^2$.

Putting together:

$c^2/(m^2 +1) = a^2$.

Answer 2).

The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if:

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