You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Equation of Circle Whose Extremities of a Diameter Given

A line cuts t...

Question

A line cuts the x-axis at A(4, 0) and the y-axis at B(0, 8). A variable line PQ is drawn perpendicular to AB cutting the x-axis in P and the y-axis in Q. If AQ and BP intersect at R, find the locus of R.

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

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

Open in App

Solution

The correct option is D

$∠ A R B = \frac{π}{2}$
The locus is equation of the circle having A, B as end points of diameter i.e $x^{2} + y^{2} - 4 x - 8 y = 0$

Suggest Corrections

Similar questions

A line cuts the x - axis at A(7, 0) and the y - axis at B(0, -5). A variable line PQ is drawn perpendicular to AB. Cutting the x - axis at P and the y - axis at Q.
If AQ and BP intersect at R, the locus of R is

Q. A line cuts the x-axis at

A (5, 0)

and the y-axis at

B (0, - 3)

. A variable line

P Q

is drawn perpendicular to

A B

cutting the x-axis at

P

and the y-axis at

Q

. If

A Q

and

B P

meet at

R

, then the locus of

R

Q. The equation of the circle which touches x-axis and whose centre is (1, 2), is

Q. A straight line

L_{1} : \frac{x}{a} + \frac{y}{b} = 1

intersects the

x

-axis and

y

-axis at

P

and

Q

respectively and a straight line

L_{2}

perpendicular to

L_{1}

cuts the

x

-axis and

y

-axis at

R

and

S

respectively. The locus of the point of intersection of the lines

P S

and

Q R

Q. The line

\frac{x}{a} + \frac{y}{b} = 1

cuts the axis at

A

and

B

,another line perpendicular to

A B

cuts the axes at

P, Q

respectively.Locus of points of intersection of

A Q

and

B P

Join BYJU'S Learning Program

A line cuts the x-axis at A(4, 0) and the y-axis at B(0, 8). A variable line PQ is drawn perpendicular to AB cutting the x-axis in P and the y-axis in Q. If AQ and BP intersect at R, find the locus of R.

The correct option is D ∠ARB=π2 The locus is equation of the circle having A, B as end points of diameter i.e x2+y2−4x−8y=0

The correct option is D

$∠ A R B = \frac{π}{2}$
The locus is equation of the circle having A, B as end points of diameter i.e $x^{2} + y^{2} - 4 x - 8 y = 0$