Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8

Q.

The differential equation of the family of circles passing through the fixed points (a, 0) and (- a, 0), is 

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

a

y1y2x2+2xy+a2=0

b

none of these

c

y1y2+xy+a2x2=0

d

y1y2x2+a2+2xy=0

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

detailed_solution_thumbnail

Let the equation of the family of circles be

x2+y2+2gx+2fy+c=0

It passes through (a, 0) and (- a, 0). Therefore, 

a2+2ag+c=0 and a22ag+c=0

Solving these two equations, we get c=a2 and g=0

Substituting the values of c and gin (i), we get

x2+y2+2fya2=0

It is a one parameter family of circles. 

Differentiating with respect to x, we get 

2x+2yy1+2fy1=0f=x+yy1y1

Substituting the value off in (ii), we get

y1y2x2+a2+2xy=0

This is the required differential equation

Watch 3-min video & get full concept clarity

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon
The differential equation of the family of circles passing through the fixed points (a, 0) and (- a, 0), is