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 eqn of the circle which passes through the points (3,-2) , (-2,0) and has the centre on the line 2x-y =3

find the eqn of the circle which passes through the points (3,-2) , (-2,0) and has the centre on the line 2x-y =3

Grade:12th pass

2 Answers

Yash Jain
55 Points
6 years ago
Lemme give you some hints.
See, for a circle, the perpendicular drawn to any of its chord passes through its center. Also, it bisects the chord. So if we find a perpendicular line which is passing through the mid-pt of (3,-2) and (-2,0) i.e., (1/2,-1). That line would be 5x-2y=9/2. On solving this equation with 2x-y=3, we’ll get the co-ordinates of the center. Center will come out as (-3/2,-6). Thus, radius is sqrt(145/4).
Hence, the final eq of the req circle is:- 4x^2 + 4y^2 + 12x + 48y + 8 = 0
Abhishek Singh
93 Points
6 years ago
 
the equation of a family of circles passing through the given pts are

(x-3)(x+2)+(y+2)y+k(5y+2x+4)=0  where k is a parameter
the centre of circle is     1-2k/2 , -(5k+2)/2
Since this lies on the given line by substituting we get k=2
hence circle is x2+y2+ 3x+12y+2=0

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