0

I am brushing up some plane and solid analytic geometry before taking a course on multivariable calculus. I am deriving important results and solving through the book, Co-ordinate Geometry by SL Loney. I am stuck on a question in circles, I feel the equation becomes unwieldy.

Examples XVIII, problem 14

Find the equation to the straight lines joining the origin to the points in which the straight line $y=mx+c$ cuts the circle

$$x^2+y^2=2ax+2by$$

Hence, find the condition that these points may subtend a right angle at the origin.

Find also the condition that the straight line may touch the circle.

Solution(My attempt).

Substituting the equation of the line $y=mx+c$ in the equation of the circle,

$\begin{aligned} x^2+y^2&=2ax+2by\\ x^2+(mx+c)^2&=2ax+2b(mx+c)\\ x^2+m^{2}x^{2}+2mcx+c^2&=2ax+2bmx+2bc\\ \implies (1+m^2)x^2+(2mc-2a-2mb)x+c^2-2bc&=0 \end{aligned}$

I am unable to progress beyond this point. Any tips, hints or suggestions would really help!

Community
  • 1
Quasar
  • 5,410
  • without checking the details, you have a quadratic which you can solve to give two solutions for $x$ and so from $y=mx+c$ two corresponding solutions for $y$. If you then have a solution $(x_1,y_1)$ then the line through the origin is $y=\frac{y_1}{x_1}x+0$ – Henry Feb 01 '19 at 12:44

1 Answers1

1

We may let $t=\frac{y}{x}$ and solve the equation about $t$: $$ x^2+y^2 = 2ax + 2by \implies 1+t^2 =\frac2{x}(a+bt), $$ $$ y=mx+c\implies t=m+\frac{c}x\implies \frac2 x = 2\frac{t-m} c. $$ Thus $$ 1+t^2 = 2\left(\frac{t-m} c\right)(a+bt), $$ which is equivalent to $$ (2b-c)t^2 +2(a-mb)t-(2am+c)=0.\tag {*} $$ The solution is given by $$ t_i=\frac{(a-mb)\pm\sqrt{(a-mb)^2+(2b-c)(2am+c)}}{2b-c},\ \ i=1,2, $$ provided that $D/4 =(a-mb)^2+(2b-c)(2am+c)=-c^2+(2b-2am)c+(a+mb)^2\ge 0$.

These two lines $y=t_i x$ form a right angle at the origin, that is, are orthogonal to each other if and only if $t_1t_2 = -1$. By Vieta's formula, this is equivalent to $\frac{2am+c}{2b-c}=1$, i.e. $b=am+c$. We note that this is also equivalent to that the line $y=mx+c$ passes through the center of the circle $(a,b)$.

The line $y=mx+c$ touches the circle if and only if the equation $(*)$ has multiple roots. i.e. $D/4 =0$. Arranging the terms, we find that it is equivalent to $$ -(c+am-b)^2 +(a^2+b^2)(m^2+1)= 0\Longleftrightarrow c= b-am\pm \sqrt{(a^2+b^2)(m^2+1)}. $$ We also note that this is equivalent to $$ \frac{|b-am-c|}{\sqrt{m^2+1}}=\sqrt{a^2+b^2}, $$ where the LHS is the distance between the line $y-mx-c=0$ and the center of the circle $(a,b)$, and the RHS is the radius of the circle.

Myunghyun Song
  • 21,723
  • 2
  • 24
  • 60