How to find the equation of a line, tangent to a circle, that passes through a given external point

Question

I am given the equation of a circle: $(x + 2)^2 + (y + 7)^2 = 25$. The radius is $5$. Center of the circle: $(-2, -7)$.

Two lines tangent to this circle pass through point $(4, -3)$, which is outside of said circle. How would I go about finding one of the equations of the lines tangent to the circle?

I haven't started calculus, so I ask for advice fitting for someone starting a topic like this.

Answer 1

A useful technique for this situation which avoids using calculus goes like this:

Find the equation of the line through the point $(4,-3)$ with gradient $k$, and then use this (linear) equation to substitute into the equation of the circle to get a quadratic. The two roots of this quadratic would give the 2 points of intersection of the line with the circle, but in our case, we want to know the values of $k$ for which the quadratic has a double root (i.e. the 2 points of intersection are actually coincident). Use the standard condition ($b^2=4ac$) for a double root of a quadratic.

In your case, the equation of the line through $(4,-3)$ with gradient $k$ is $y+3 = k(x-4)$, or $y+7 = k(x-4)+4$. Substituting this into the equation of the circle gives:

$$(x+2)^2 + (k(x-4)+4)^2 = 25$$

So you need to rearrange this quadratic in $x$ and find out which values of $k$ give double roots.

Answer 2

One possible approach would be the following:
1) Denote by $A$ the point $(4,-3)$, by $O$ the center of the circle, and by $P_1,P_2$ the two points on the circle in which the corresponding tangent line intersects the circle. Then both the triangles $AOP_1$ and $AOP_2$ are right, with hypotenuse $AO$.
2) The length of $AO$ is $\sqrt{(-2-4)^2+(-7-(-3))^2}=\sqrt{36+16}=\sqrt{52}$. The length of $OP_1$ and of $OP_2$ is the radius, and therefore equal to $5$. So, using the Pythagorean theorem, we get: $$|AP_1|=|AP_2|=\sqrt{|OA|^2-r^2}=\sqrt{52-25}=\sqrt{27}$$ 3) From here you can find both $P_1$ and $P_2$ - they are the points of intersections of the given circle with the circle of radius $\sqrt{27}$ around $(4,-3)$.
4) Now find the equations of two lines passing through given points.

Answer 3

Assume the point where the tangent touches the circle is $(x_1,y_2)$, then this point satisfies the equation of the circle and the equation

$$ y'= \frac{y_1+3}{x_1-4 }\longrightarrow (*).$$

Now, you need to find $y'$ and evaluate it at the point $(x_1,y_1)$ and subs. back in (*), so you end up having two Equations in two variables. Solve them to find the point $(x_1,y_1)$ and finaly you will have two points that required to find the tangent plane.

Answer 4

The equation of any line passing through $(4,-3)$ is $$\frac{y+3}{x-4}=m\iff mx-y-4m-3=0$$ where $m$ is the gradient

Now, the perpendicular distance of tangent to center of circle = radius of circle

Do you know how to calculate the perpendicular distance of a line from the point?

Observe that the values of $m$

$(i)$ will be distinct real (hence, two tangents) if the point lies outside the circle,

$(ii)$ will be equal real (hence, one tangent) if the point lies on the circle,

$(iii)$ will be distinct imaginary (hence, no real tangents) if the point lies inside the circle,

See $12(b)$ here