Tangents from a certain point, to a circle?

Question

We have a circle with radius 2, centred on the origin. Find the equation of the lines passing through the point $(0,4)$ which are tangent to the circle.

So we have the circle $$x^2 + y^2 = 4$$

We need 2 lines $$y=ax+4$$

So if you fill that in the equation of the circle, we end up with $$a^2x^2 + 8ax + x^2 + 12 = 0 $$.

So we'd have to solve $$ a = \dfrac{-8x \pm \sqrt{64x^2-4x^2(x^2+12)}}{2x^2}$$

I must have done something wrong, since this causes the discriminant to be negative, and thus gives no answer. Is there a smarter way I overlooked?

Answer 1

Suppose $A(x_1,y_1)$ is the point of the circle where the tangent is drawn.

Then the equation of the tangent is $xx_1+yy_1=4$. Since $(0,4)$ belongs to the tangent it satisfies her equation so $0x_1+4y_1=4\rightarrow y_1=1$.

Then subtitute this value to the circle equation

we get $x_1^2+y_1^2=4\rightarrow x_1^2+1^2=4\rightarrow x_1=\sqrt3$ or $x_1=-\sqrt3$.

Thus we have two tangents with equations:

$\sqrt3x+y=4$ and $-\sqrt3x+y=4$.

Answer 2

The values of $x$ represents the abscissa of the intersection

The equation will be $$(a^2+1)x^2+8ax+12=0$$

For tangency, the roots of the above quadratic Equation must be same, hence the discriminant must be $0$

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Alternatively,

The equation of any line passing through $(0,4)$ will be $\frac{y-4}{x-0}=m\iff mx-y+4=0 $ where $m$ is the gradient

Now, the perpendicular distance of a tangent from the center $(0,0)$ of the circle will be $=$ radius$(=2)$

Answer 3

You can get the tangent points by doing the intersection between the circle and the circle of radius $2$ centered in $(2,0)$ (the lines from one of this points to $(0,0)$ and from the same point to $(4,0)$ will be perpendicular)

so you have

$$x^2 + y^2 = 4$$

and

$$(x-2)^2 + y^2 = 4$$

The x coordinate will be the same for the two points, so $$x^2 - 4 = (x-2)^2 -4$$ from here you get $x=1$ and $y^2=3$

So you have the two points of tangency, and from here you can easily find the equations of the tangents.