Finding locus of centre of circle

Question

The locus of the center of the circles such that the point $(2,3)$ is midpoint of the chord $5x +2y =16$ is

(A) $2x-5y+11=0$

(B) $2x+5y-11=0$

(C) $2x+5y+11=0$

(D) None of these

I do not want solution, the real question is given below

This is how I solved it:

Let, the center be (h,k),now the distance between (2,3) and (h,k) equals to distance of (h,k) from line $5x +2y =16$, thus $$(h-2)^2+(k-3)^2 = \frac{(5(h)+2(k)-16)^2}{5^2+2^2}$$ You can see I have already squared both sides. Now this clearly gives a 2 degree equation not a linear one as given in options.

But when I saw solution of this question, its given that $$(\frac{-5}{2})(\frac{k-3}{h-2})=-1$$Now this gives option A, .i.e. a linear locus, So you can see that I have a 2 degree locus, but the book has 1 degree, how can both be correct at the same time, is any one of the solutions wrong, is so explain, and if both are correct, then explain why there are 2 different locus for same conditions.

Answer 1

The center of the circle is on the perpendicular $(p)$ in $(2,3)$ on the line with equation $5x+2y=16$, having slope $m=-\frac 52$, so $(p)$ has slope $m'=-\frac 1m=\frac 25$, so it has an equation of the shape $5y-2x=?$, and because $(2,3)$ satisfies, the equation is $$ 5y-2x=11\ . $$ We get the solution (A).

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Now consider also the equation of degree two... $$ (x-2)^2+(y-3)^2 = \frac{(5x+2y-16)^2}{5^2+2^2}\ . $$ We can equivalently rewrite it successively: $$ \begin{aligned} (5^2+2^2)((x-2)^2+(y-3)^2) &= (5x+2y-16)^2\ ,\\ (5^2+2^2)((x-2)^2+(y-3)^2) &= (\ 5(x-2)+2(y-3)\ )^2\ ,\\ 5^2(x-2)^2+5^2(y-3)^2+2^2(x-2)^2+2^2(y-3)^2 &= 5^2(x-2)^2+2\cdot5(x-2)\cdot 2(y-3)+2^2(y-3)^2\ ,\\ 5^2(y-3)^2+2^2(x-2)^2 &= 2\cdot5(x-2)\cdot 2(y-3)\ ,\\ (\ 5(y-3)-2(x-2)\ )^2 &=0\ . \end{aligned} $$ This is also (A), over a field.

Answer 2

The quadratic equation appears when you square the expressions.

Geometrically, a given distance to the point gives a circle and the same distance to the line gives two lines, on both sides, so that for a given common distance there are two solutions. In fact, your equation is a perfect square, representing twice the same line.

We see this by moving the point to the origin and writing

$$h^2+k^2=\frac{(ah+bk)^2}{a^2+b^2}.$$

It is easy to check that this is also

$$\frac{(bh-ak)^2}{a^2+b^2}=0.$$

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You could have solved using the algebraic distance to the line (i.e. positive on one side, negative on the other), and process

$$\sqrt{h^2+k^2}=\frac{ah+bk}{\sqrt{a^2+b^2}},$$

(without an absolute value).

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At the same time this gives the solution, as the linear coefficients must be proportional to $(b,-a)$.

Answer 3

Your conclusion that since the equations are of differing degree there are two different loci is incorrect. What you have are two equations for the same set of points. The second-degree equation that you derived simplifies to $$\frac1{29}(5k-2h-11)^2 = 0.$$ The given solution, on the other hand, simplifies to $\frac12(5k-2h-11)=0$. It should be obvious that the graphs of the two equations are identical.