# Equation of a tangent line to a given circle from an external point - Stuck!

by JOhnJDC
Tags: circle, equation, point, quadratic, tangent
 P: 40 Sorry, accidentally hit submit before I was finished. Anyway.. Your attempt is good, but doesn't use the fact that the lines are tangent to the circle. First you'll want the center of the circle. \begin{align*} x^2 + y^2 &= 2y\\ x^2 + y^2 - 2y &= 0 \end{align*} Complete the square (I'll leave the details to you) to get: $$x^2 + (y-1)^2 = 1$$ So the center of the circle is (0, 1), and the radius is 1. You're right about the equation of the tangent line. But what you also need to use is that the tangent line to a circle at point P is perpendicular to the line that goes through point P and the center of the circle. So you want to find the equation of all lines that go through (0, 4) and are tangent to the circle. Such a line will hit the circle at some point (a, b). So you want a line that goes through the points (0,4) and (a,b). This line is perpendicular to the line that goes through (a, b) and (0, 1), since (0, 1) is the center of the circle. The line through the points (0,4) and (a,b) has slope m = (b-4)/a. (That's the line you want to find.) The line through the points (0,1) and (a,b) has slope n = (b-1)/a. Since these lines are perpendicular, their slopes are negative reciprocals. So you have m = -1/n, which means (after some simplifications): $$\frac{b-4}{a} = \frac{a}{1-b}$$ Multiply both sides by a*(1-b), and you get $$a^2 = (b-4)(1-b)$$ Note that the point (a,b) is on the circle, so it satisfies the equation of the circle. In other words, we have \begin{align*} a^2 + (b-1)^2 &= 1\\ a^2 &= 1 - (b-1)^2 \end{align*} So now we have two different expressions for a^2: $$$a^2 = (1-b)(b-4)$ $a^2 = 1 - (b-1)^2$$$ So you want to set them equal to each other and solve for b: \begin{align*} (1-b)(b-4) &= 1 - (b-1)^2\\ b - 4 - b^2 + 4b &= 1 - (b^2 - 2b + 1)\\ &\vdots \end{align*} I'll leave the details to you, but you should end up with b = 4/3. Then you plug that back into either one of your equations for a^2, and you'll get $$a = \pm \frac{2\sqrt{2}}{3}$$ Then you can plug these into one of your expressions for m, and you'll get something like \begin{align*} m &= \frac{a}{1-b}\\ &= \frac{\pm\frac{2\sqrt{2}}{3}}{1 - \frac{4}{3}}\\ &= \mp 2\sqrt{2} \end{align*} ("m" has the oppposite sign of "a".. that is, if one is positive, the other is negative) So the equations of your tangent lines are $$y = \pm2\sqrt{2}x + 4$$