Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

find the length of the chord joining the points in which the straight line, (x/a)+(y/b)=1, meets the circle x^2+y^2=r^2

find the length of the chord joining the points in which the straight line, (x/a)+(y/b)=1, meets the circle x^2+y^2=r^2

Grade:12

2 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago

Hi Suhas,

 

Consider (p,q) to be the midpoint of the chord.

So T=S1 is its equation.

which is px+qy = p2+q2---------(this eqn is the same as x/a+y/b=1)

So ap = bq = ??p2+q2.

Let the line intersect the cirlcea at (x1,y1) and (x2,y2)

So D = √[ (x1-x2)2+(y1-y2)2 ] = √[ (x1+x2)2+(y1+y2)2 - 4x1x2 - 4y1y2 ]

Now x1+x2 = 2p, y1+y2 = 2q --------(As (p,q) is the midpoint).

 

From the line equation px+qy = p2+q2, eliminate y from the circle eqn. You'll get a QE in x, for which x1,x2 are the roots.

So you'll get product of roots = x1x2 = p2+q2 - r2q2.

similarly y1y2 = p2+q2 - r2p2.

 

Substitute in the distance, and you'll get

D = √[4r2 - 4(p2+q2)]

 

Now ap=bq=p2+q2.

Also p/a+q/b=1

So p2+q2 = (a2b2)/(a2+b2)......

 

Now substitute and get D, interms of r,a,b

 

Best Regards,

Ashwin (IIT Madras).

User
5 Points
2 years ago
From the circle centre (0,0), drop a perpendicular to the line, bx+ay-ab=0. Its length, p=ab/√(a²+b²)
The chord length, L = 2[√(r² -p²)] = 2√[(r² -(ab)²/(a²+b²))].

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free