3
$\begingroup$

Prove that the locus $|z-z_1|^2 +|z-z_2|^2=k$ of a moving point $z$ on the Argand plane is a circle when $|z_1-z_2|^2\leqslant 2k$.

What i have tried out- $|z-1|^2+|z-2|^2=k$ let $z=x+iy$ $$\\ \implies2x^2-6x+5+2y^2-k=0$$ This is the equation of a circle. centre $=(\frac32,0)$ radius=$\sqrt{\dfrac{(2k−1)}4 }$ means $\\ k\geqslant\frac12$

$\endgroup$
2
$\begingroup$

Try with $z=x+iy, z_1=a+ib, z_2=c+id$

On rearrangement we have, $$x^2+y^2-x(a+c)-y(b+d)+\frac{a^2+b^2+c^2+d^2-k}2=0$$

$$\implies \left(x-\frac{a+c}2\right)^2+\left(y-\frac{b+d}2\right)^2=\frac{2k-2a^2-2b^2-2c^2-2d^2+(a+c)^2+(b+d)^2}4=\frac{2k-\{(a-c)^2+(b-d)^2\}}4$$

For real circle we need $2k-\{(a-c)^2+(b-d)^2\}\ge0$

But here, $|z_1-z_2|=?$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.