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.
$\endgroup$ 47 Answers
$\begingroup$A useful technique for this situation which avoids using calculus goes like this:
Find the equation of a 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 vales of $k$ give double roots.
$\endgroup$ 1 $\begingroup$Without any loss of generality we can draw a tangent to a circle with centre $C\equiv(C_x,C_y)$ and radius $r$ and choose a point$P\equiv(P_x,P_y)$ on the tangent so that its distance from the centre is$\overline{CP}\equiv d$ with $d \ge r$.
From the sketch above we recognize that the tangency points can be described as the sums of vectors, i.e.,
$$ \overrightarrow{OT_1} = \overrightarrow{OC}+ \frac{a}{d}\,\overrightarrow{CP}+ \frac{b}{d}\,\overrightarrow{CP}',\qquad \overrightarrow{OT_2} = \overrightarrow{OC}+ \frac{a}{d}\,\overrightarrow{CP}- \frac{b}{d}\,\overrightarrow{CP}'$$
where the prime in $\overrightarrow{CP}'$ represents a 90⁰ anticlockwise rotation of the vector.
Because our task is now to determine $a/d$ and $b/d$, we observe that the triangles $CT_1C'$ and $CPT_1$ are similar so that, with the position$\rho=r/d$, we have
$$\frac{a}{r} = \frac{r}{d} \implies \frac{a}{d}=\left(\frac{r}{d}\right)^2 =\rho^2$$
and
$$\frac{b}{r} = \frac{R}{d} \implies \frac{b}{d}=\frac{rR}{d^2} =\rho\frac Rd$$
but $R=\sqrt{d^2-r^2}$ and eventually we can write
$$ \frac{b}{d}= \rho\sqrt{1-\rho^2}.$$
From the last relation we recognize 1) that a real solution requires$\rho\le1$ (i.e., $P$ is not inside the circle) and 2) that for$\rho=1$ the point $P$ is on the circumference and the tangency points degenerate into $P$.
The OP can now compute the tangency points, etc.
E.g., in Python 3 (that is as close to pseudo-code as it could possibly be...)
from math import sqrt
# Data Section, change as you need #
Cx, Cy = -2, -7 #
r = 5 #
Px, Py = 4, -3 #
# ################################ #
dx, dy = Px-Cx, Py-Cy
dxr, dyr = -dy, dx
d = sqrt(dx**2+dy**2)
if d >= r : rho = r/d ad = rho**2 bd = rho*sqrt(1-rho**2) T1x = Cx + ad*dx + bd*dxr T1y = Cy + ad*dy + bd*dyr T2x = Cx + ad*dx - bd*dxr T2y = Cy + ad*dy - bd*dyr print('The tangent points:') print('\tT1≡(%g,%g), T2≡(%g,%g).'%(T1x, T1y, T2x, T2y)) if (d/r-1) < 1E-8: print('P is on the circumference') else: print('The equations of the lines P-T1 and P-T2:') print('\t%+g·y%+g·x%+g = 0'%(T1x-Px, Py-T1y, T1y*Px-T1x*Py)) print('\t%+g·y%+g·x%+g = 0'%(T2x-Px, Py-T2y, T2y*Px-T2x*Py))
else: print('''\
Point P≡(%g,%g) is inside the circle with centre C≡(%g,%g) and radius r=%g.
No tangent is possible...''' % (Px, Py, Cx, Cy, r))Executing the program yields
The tangent points: T1≡(-1.1139,-2.07914), T2≡(2.88314,-8.0747).
The equations of the lines P-T1 and P-T2: -5.1139·y-0.920857·x-11.6583 = 0 -1.11686·y+5.0747·x-23.6494 = 0One 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.
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.
$\endgroup$ $\begingroup$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
$\endgroup$ $\begingroup$My work out
PC gradient:
(-3+7)/(4+2)=tan(33.7°)
PC distance:
√((-3+7)²+(4+2)²)=√52
∠CPQ = arcsin(5/√52)=43.9°
Equations of tangent passing through PQ & PR
(Y+3)/(X-4)=tan(33.7±43.9)
Let eq of tangents be y= mx+ c then it passes through point (4,-3)
So -3=4m+c let it put aside
Now tangent touches circle so radius = distance between centre and line
5 = |-2m+7+c|/(m^2+1)^1/2
By solving we get
11 m^2 +24m=9
We get two values of m
These are slopes of linesenter preformatted text hereWe can equation of lines and angle between lines
This is my idea
