The locus of a point which moves such that the line segment having end points $(2, 0)$ and $(- 2, 0)$ subtend a right angle at that point

x + y = 4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x^{2} + y^{2} = 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} - y^{2} = 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n o n e o f t h e s e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $x + y = 4$
Let C be a point on the locus
$A P ⊥ B P$
Gradient of $A P$
$(m_{A C}) = \frac{y - 0}{x - 2} = \frac{y}{x - 2}$
Gradient of $A P$
$(m_{B C}) = \frac{y - 0}{x - (- 2)} = \frac{y}{x + 2}$
For $A P ⊥ B P$
$(m_{A P}) \times (m_{A C}) = - 1$
$\frac{y}{x - 2} \times \frac{y}{x + 2} = - 1$
$y^{2} = - (x^{2} - 2^{2})$
$x^{2} + y^{2} = 4$
Hence, this is the required solution.

Suggest Corrections

Similar questions

Find the locus of a point such that the line segments having end points(2,0) and (-2,0) subtend a right angle at that point.

x^{2} + y^{2} = 4

x^{2} = 2 (x + y)

x^{2} + y^{2} - 2 x - 2 y = 0

A point moves in such a way that the sum of square of its distance from the points A(2, 0) and B(-2, 0) is always

equal to the square of the distance between A and B. The locus of the point is

The locus of the middle points of those chords of the circle $x^{2} + y^{2} = 4$ which subtend a right angle at the origin is

(x, y)

\sqrt{(x - 2)^{2} + y^{2}} + \sqrt{(x - 2)^{2} + y^{2}} = 4

(- 2, 0)

(2, 0)

4.

Join BYJU'S Learning Program