Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
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
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).
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !