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If the chord $x+y=b$ of the curve $x^2+y^2-2ax-4a^2=0$ subtends a right angle at the origin, prove that: $b(b-a)=4a^2$

My Approach.

Given,

Equation of the chord, $$x+y=b$$ $$\frac {x+y}{b}=1$$

Now,

Equation of the curve, $$x^2+y^2-2ax-4a^2=0$$ $$x^2+y^2-2ax=4a^2$$ $$(b-y)^2+(b-x)^2-2ax=4a^2$$

I got stuck at here. Please help me to complete it.

pi-π
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  • What does "subtends...whatever... at the origin " mean?? I can understand that if "we look" at the cord from any point on the circle then "subtending an angle at that point" means the inscribed angle determined by that point and that cord. Yet the origin is not a point on the circle... – DonAntonio Sep 04 '16 at 10:46
  • @DonAntonio, I, too could not understand the question. However, the question is absolutely correct as given. – pi-π Sep 04 '16 at 10:51
  • Then it must be me not acquainted enough with the terms used here. – DonAntonio Sep 04 '16 at 10:53
  • @DonAntonio, It might be. I am too, unclear with the question. – pi-π Sep 04 '16 at 10:58
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    @DonAntonio it just means the lines from the origin to the endpoints of the chord are perpendicular. – David Quinn Sep 04 '16 at 11:02
  • @DavidQuinn Thanks, that sounds plausible...but never met it before. – DonAntonio Sep 04 '16 at 11:03

2 Answers2

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First solve the equations simultaneously, and we arrive at the quadratic equation $$2x^2-2x(a+b)+b^2-4a^2=0$$

The roots satisfy $$x_1+x_2=a+b$$ and $$x_1x_2=\frac{b^2-4a^2}{2}$$

The perpendicularity condition can be written as $$x_1x_2+y_1y_2=0$$

$$\implies x_1x_2+(b-x_1)(b-x_2)=0$$ $$\implies 2x_1x_2-b(x_1+x_2)+b^2=0$$

Using the above results for the sum and product of the roots, the result follows immediately.

David Quinn
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1

The combined equation of the lines joining the origin to the end points of the chord can be obtained by "homogenising" the equation of the curve. This is \begin{align*} x^2+y^2 - 2ax\left(\frac{x+y}{b}\right) - 4a^2\left(\frac{x+y}{b}\right)^2 = 0 \end{align*} These lines are perpendicular if the sum of the coefficients of $x^2$ and $y^2$ is zero. Thus the required condition is \begin{align*} 1+1-\frac{2a}{b}-\frac{4a^2}{b^2}(1+1) &= 0\\ b^2 - ab -4a^2 &= 0\\ b(b-a) &= 4a^2 \end{align*}

  • What is meant by homogenising? – pi-π Sep 04 '16 at 11:09
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    @user354073 If we have an equation of a curve as $ax^2+2hxy+by^2+2gx+2fy+c = 0$ and a line $lx+my=1$, the equation of the pair of lines joining the origin and the end points of $lx+my=1$ on the curve can be obtained as $ax^2+2hxy+by^2 +2(gx+fy)(lx+my) + c(lx+my)^2 = 0$. Note that we have "created" a homogeneous equation of second degree from the curve and the line and hence it represents a pair of lines through the origin. It is easy to see that the points of intersection of the curve and line lie on this. This process is called "homogenising". –  Sep 04 '16 at 11:15
  • Also for a pair of lines $Ax^2+2Hxy+By^2 = 0$ through the origin, they are perpendicular if $A+B=0$. –  Sep 04 '16 at 11:16
  • It means, homogenising is to be done in the last three terms of a general equation of the curve. Is it? – pi-π Sep 04 '16 at 11:19
  • The aim is to make each term of second degree. So we need to multiply linear terms by $lx+my$ and the constant terms by $(lx+my)^2$. –  Sep 04 '16 at 11:24
  • How did you do $(1+1)$ while writing the perpendicularity condition? – pi-π Sep 04 '16 at 11:28
  • Coefficient of $x^2$ is 1, coefficient of $y^2$ is 1 in the first two terms, in the third term, only $x^2$ is present with coefficient $-2a/b$ and in the last term we have $x^2$ term with coefficient $-4a^2/b^2$ and $y^2$ term also has the same coefficient. We add all these and equate to zero. –  Sep 04 '16 at 11:35
  • Ahh, Thanks a lot. It was nice talking to you. – pi-π Sep 04 '16 at 11:40