MyQuestionIcon

1

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

Introduction to LOCI

From point ...

Question

From point $(3, 4)$ , chords are drawn to the circle $x^{2} + y^{2} - 4 x = 0$ . The locus of the midpoints of the chords is

A

x^{2} + y^{2} - 5 x - 4 y + 6 = 0

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

B

x^{2} + y^{2} + 5 x - 4 y + 6 = 0

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

C

x^{2} + y^{2} - 5 x + 4 y + 6 = 0

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

D

x^{2} + y^{2} - 5 x - 4 y - 6 = 0

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

Open in App

Solution

The correct option is A $x^{2} + y^{2} - 5 x - 4 y + 6 = 0$
Centre of the circle $= (2, 0)$

Let the locus be $(h, k)$
Slope of $O M = m_{1}$
Slope of $P M = m_{2}$
$\Rightarrow m_{1} m_{2} = - 1$
$\Rightarrow (\frac{4 - k}{3 - h}) (\frac{k - 0}{h - 2}) = - 1$
$\Rightarrow h^{2} + k^{2} - 5 h - 4 k + 6 = 0$
Hence, locus is $x^{2} + y^{2} - 5 x - 4 y + 6 = 0$

Suggest Corrections

thumbs-up

0

Similar questions

Q. The radical axis of circles

x^{2} + y^{2} + 3 x + 4 y - 5 = 0

x^{2} + y^{2} - 5 x + 5 y - 6 = 0

Q. The slope of the radical axis of the circles

x^{2} + y^{2} + 3 x + 4 y - 5 = 0

x^{2} + y^{2} - 5 x + 5 y - 6 = 0

Q. The locus of centres of the circles which cut the circles

x^{2} + y^{2} + 4 x - 6 y + 9 = 0

x^{2} + y^{2} - 5 x + 4 y + 2 = 0

orthogonally is

Q. The distance of

(1, - 2)

from the common chord of

x^{2} + y^{2} - 5 x + 4 y - 2 = 0

x^{2} + y^{2} - 2 x + 8 y + 3 = 0

Q. The locus of the centres of the circles which cut the circles

x^{2} + y^{2} + 4 x - 6 y + 9 = 0

x^{2} + y^{2} - 5 x + 4 y - 2 = 0

orthogonally is

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Introduction to LOCI

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Introduction to LOCI

Standard X Mathematics

Join BYJU'S Learning Program

CrossIcon