Question

The locus of a point which moves such that the line
segments having end points (2, 0) and (-2, 0) subtend a
right angle at that point
(a)x+y=4
x² + x = 4
(c)x2-y2=4
(d) none of these

Answer 1

Answer:

Option (b) $x^2+y^2=4$

Step-by-step explanation:

Let the coordinate of the point be $(x,y)$

Slope of line joining points $(2,0)$ and $(x,y)$

$m_1=\frac{y-0}{x-2}$

$\implies m_1=\frac{y}{x-2}$

Slope of line joining points $(-2,0)$ and $(x,y)$

$m_2=\frac{y-0}{x-(-2)}$

$\implies m_2=\frac{y}{x+2}$

Since both these lines are perpendicular

Therefore,

$m_1m_2=-1$

$\implies \frac{y}{x-2}\times \frac{y}{x+2}=-1$

$\implies \frac{y^2}{(x-2)(x+2)}=-1$

$\implies \frac{y^2}{x^2-2^2}=-1$

$\implies y^2=-x^2+4$

$\implies x^2+y^2=4$

Which is the locus of the point $(x,y)$

It is clear that the locus is a circle.

Hope this answer is helpful.

Answer 2

$\huge\bf{Answer:-}$

Refer the attachment.