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
Grade: 12

                        

Find the inverse point of (-2,3) with respect to the circle x^2+y^2-4x-6y+9=0

Find the inverse point of (-2,3) with respect to the circle x^2+y^2-4x-6y+9=0

4 months ago

Answers : (1)

Ram Kushwah
106 Points
							
 
 
The equation of the circle is
x²+y²-4x-6y+9=0
This can be written as
(x-2)²+(y-3)²=4=2²
      so centre of circle is O (2 3) and radius r =2
Now let a point P(h,k ) is a inverse of poin A (-2,3) w.r.t. circle
 
x²+y²-4x-6y+9=0
 
So OP x OA=r²=4
 
Now OA=√{ (2+2)²+(3-3)²}=√4²=4
Thus OP x 4=4
OP=1
OP²=1......................(1)
 
Now equation of OA is:
y-3=(3-3)/(-2-2) *(x-2)
y-3=0
y=3
P (h,k) lies on OP
so k=3
From(1)
 
OP²=1
(h-2)²+(k-3)²=1
(h-2)²+(3-3)²=1
(h-2)²=1
h-2=±1
h=3 or 1
Thus the points are (1,3 ) or (3,3)
But point P is same side of point A that is possible only(1,3)
Thus point P is (1,3)
 
 
 
29 days ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution


Course Features

  • 53 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions


Ask Experts

Have any Question? Ask Experts
 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details