Finding the locus of the midpoint of chord that subtends a right angle at $(\alpha,\beta)$

Question

There is a circle $x^2+y^2=a^2$. On any line that cuts the circle in two distinct points(it is a secant), the points of intersection with circle are taken and at those two points I draw the tangents that intersect at some point, say $(\alpha,\beta)$. It's given that the tangents intersect at right angles at the point $(\alpha,\beta)$. I need to find the locus of the midpoint of the chord.

THE BOOK'S WAY:
Let the midpoint be $(h,k)$.
Using $T=S_1$ for the equation of chord, the chord is $hx+ky=h^2+k^2$
If the chord intersects the circle at $(x_1,y_1)$ and $(x_2,y_2)$, then by condition of perpendicularity $$m_1m_2=-1$$ $$({y_1-\beta \over x_1-\alpha})({y_2-\beta \over x_2-\alpha})=-1$$ $$(x_1x_2+y_1y_2)+(\alpha^2+\beta^2)=\alpha(x_1+x_2)+\beta(y_1+y_2)$$ Then using the equation of chord and separately eliminating $x,y$ we obtain quadratics $$\lambda x^2-2\lambda hx+\lambda^2-a^2k^2=0$$ $$\lambda y^2-2\lambda ky+\lambda^2-a^2h^2=0$$ where $\lambda=h^2+k^2$. Using the values of product of roots and sum of roots, the locus is $$x^2+y^2-\alpha x-\beta y+{1 \over 2}(\alpha ^2+\beta ^2-a^2)=0$$ MY WAY:
I translate the origin to $(\alpha,\beta)$ so that $x=X+\alpha$, $y=Y+\beta$
Circle becomes $X^2+Y^2+2\alpha X+2\beta Y+\alpha^2+\beta^2-a^2=0$
To get the equation of pair of lines that join the circle and chord intersection points to origin(translated), I homogenise the equation of circle, then retranslate the axes to previous origin by using $X=x-\alpha$, $Y=y-\beta$ and then put the sum of coefficent of $x^2$ and coefficient of $y^2$ equal to zero. Then I get an equation that doesn't match the book's answer. What's the flaw? Thanks in advance[Sorry if that's a long question]

Answer 1

Spotted the mistake, sorry for assurance before.

You can't put $X=x-\alpha,Y=y-\beta$ before adding the leading coefficients to zero .

1)Using transforming back, your curve is no longer a homogenized curve .

2)That pair of lines will not for sure subtend a right angle at the origin formed by $(x,y)$ system . bcoz they're meant for $(X,Y)$ system and subtend right angle at the origin of $(X,Y)$ system .

Just add the leading coefficients to zero before transforming back and then aply transformations on $(h,k)$ which are also in $(X,Y)$ system.

Answer 2

You must be knowing that the director circle subtends right angle tangents to the the circle. Here the director circle for [x^2 + y^2 = a^2] is [x^2 + y^2 = 2a^2] (you may prove it by simple geometry)

1.From an arbitary point on DC, P(a*sqrt2*cos(w),a*sqrt2*sin(w)) make a Chord Of Contact on original circle as T=0 for P.

2.Take (h,k) as the mid point of the chord on the original circle, we can form the of the as T=S1

• from 1 and 2 we get 2 independently formed eqns of the same chord therefore they are representing the equation.thus the equations are coincident so compare them by ratio of x,y and constants equal.
• We would get three relations with h,k,cos and sine. eliminate cos and sine by squaring the sum equal to 1
• you would get an equation in h and k simplify and you'll get h^2 + k^2 = (a^2)/2 that is a circle with radius as Sin(45)*a

Sorry for poor roots and inability to type roots. w is theta in P coordinates

PS: are you preparing for jee?

Answer 3

Parameterize circle as $(a \cos \alpha, a \sin \alpha)$ with $\alpha \in \left[ 0 , 2 \pi \right]$, now the tangent vector is given as $(-a \sin \alpha , a \cos \alpha)$. Let the two points correspond to end of chord be at angle $\alpha$ and at angle $\beta$, then:

$$ (a \cos \alpha, a \sin \alpha) \cdot ( a \cos \beta , a \sin \beta)=0$$

Since the tangents intersect at ninty degrees, then:

$$ \cos (\alpha - \beta)=0$$

WLOG, let $\alpha>\beta$ then $\alpha - \beta= \frac{\pi}{2}$ or $ \alpha - \beta = \frac{3 \pi}{2}$ the position vector for mid point of the chord is given as:

$$ (X,Y)=( a \frac{ \cos \alpha + \cos \beta}{2} , a \frac{ \sin \alpha + \sin \beta} {2})=(a \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha-\beta}{2},a \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2})$$

Taking the modulus:

$$ X^2 +Y^2 = a^2 \cos^2 \frac{\alpha - \beta}{2}$$

But, $\alpha - \beta= \frac{\pi}{2}$ or $ \alpha - \beta = \frac{3\pi}{2}$, hence: $ \cos^2 \alpha - \beta = \frac{1}{2}$, simplfying our equation to:

$$ X^2 +Y^2 = \frac{a^2}{2}$$

QED

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More generally locus of midpoint of chord, between point making angle $\alpha$ and $\beta$ is given as:

$$ X^2 +Y^2 = a^2 \cos^2 ( \frac{\alpha - \beta}{2})$$