# Thread: Lines tangent to a circle

1. ## Lines tangent to a circle

1. The problem statement, all variables and given/known data
You have a circle with the equation x + (y + 1) = 1. You can draw to two tangent lines to that circle that intersect the point (0,1) What are the equations of these lines? And you can't use any calculus, derivatives and the like.

2. Relevant equations
y=mx+b
x + (y + 1) = 1

3. The attempt at a solution
Well you can begin by knowing that the lines y-int will be 1 so y=mx+1. Then you can solve the equation for the circle for y which gives you y = -1 . Then you can set that equation equal to 0 and get (after factoring) . Then you can plug this into the quadratic formula to get your x, but there I get stuck. i try to plug that back into y = mx +1 but I don't know what I am looking for.

2. well, you can do this..

Let $(a,b)$ be a point on the circle $x^2 + (y+1)^2 = 1$ such that the line connecting $(a,b)$ and $(0,1)$ is tangent to the circle.

so, using two-point form of a line, you get $y - 1 = \frac{b-1}{a} \, x$ (line) with slope $\frac{b-1}{a}$ (slope 1)

note that if a line $l$ is tangent to a circle, then the line connecting the center of the circle and the point of tangency is perpendicular to $l$.

the center of the circle if $(0,-1)$. so the slope of the line from this center of $(a,b)$ will be $\frac{b+1}{a}$.

but this slope should be equal to the negative reciprocal of the slope of slope 1.

thus, $\frac{b+1}{a} = - \, \frac{a}{b-1} \Longleftrightarrow b^2-1=-a^2 \Longleftrightarrow a^2 + b^2 - 1 = 0$.. -- (1)

(a,b) is on the circle, thus, it satisfies the equation of the circle..
$a^2 + (b+1)^2 = 1 \Longleftrightarrow a^2 + b^2 + 2b = 0$ --(2)

and you have 2 equations and 2 unknowns...

you will get 2 ordered pairs and substitute it on the equation line to get the equation of the two lines..

3. Hello,

I suggest two ways.

(1) (Cheating?) Use formulae.
"Given a circle C: (x-a)^2+(y-b)^2=r^2 and a point P(X, Y), what is the line L: (X-a)(x-a)+(Y-b)(y-b)=r^2?
(i) If P is on C, L is the line tangent to C at P.
(ii) If P is outside C, let M and N be the line tangent to C passing through P. If M meets C at the point A and N meets C at the point B, L is the line AB."
In this case, by (ii), 0x+(1+1)(y+1)=1 is the line joining the tangential points. x^2+(y+1)^2=1 and 2(y+1)=1 together gives the coordinates of the tangential points. Using (i), you can deduce the tangential lines at these points.

(2) (Better answer...) x^2+(y+1)^2=1 and y=mx+1 together gives the coordinates of the points belonging to both the circle and the line. To say that the line is tangent to the circle means that the intersection consists of only 1 point. Thus, the quadratic equation x^2+(mx+1+1)^2=1 or (m^2+1)x^2+4mx+3=0 has only 1 solution. The discriminant D is D/4=(2m)^2-3(m^2+1)=0, and you get m.

Bye.

4. Originally Posted by wisterville
Hello,

I suggest two ways.

(1) (Cheating?) Use formulae.
"Given a circle C: (x-a)^2+(y-b)^2=r^2 and a point P(X, Y), what is the line L: (X-a)(x-a)+(Y-b)(y-b)=r^2?
(i) If P is on C, L is the line tangent to C at P.
(ii) If P is outside C, let M and N be the line tangent to C passing through P. If M meets C at the point A and N meets C at the point B, L is the line AB."
In this case, by (ii), 0x+(1+1)(y+1)=1 is the line joining the tangential points. x^2+(y+1)^2=1 and 2(y+1)=1 together gives the coordinates of the tangential points. Using (i), you can deduce the tangential lines at these points.
i was thinking of doing this solution too, however, since the poster said not to use calculus and derivatives, i thought using "algebraic geometry" approach, won't be good for him..

anyways.. this is good!

5. You want the line $mx-y+1=0$ to be tangent to the circle of radius 1 centred at (0,-1). That is the same as saying that the distance from (0,-1) to the line should be 1. But the distance d from the point (p,q) to the line ax+by+c=0 is given by the formula $d = \frac{|ap+bq+c|}{\sqrt{a^2+b^2}}.$

Apply that formula, and you get $1 = \frac2{\sqrt{m^2+1}}$, from which $m=\pm\sqrt3$.