Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

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

the locus of centres of all circles which touch the line x=2a and cut the circle x^2+y^2=a^2 orthogonally is


2 years ago

Deepak Kumar Shringi
4402 Points


2 years ago
Samyak Jain
333 Points
							Let equation of circle be x2 + y2 + 2gx + 2fy + c = 0,where centre is (– g,– f) and radius is $\dpi{100} \sqrt{g^{2} + f^{2} - c}$x – 2a = 0 is a tangent to the circle.$\dpi{80} \therefore$ perpendicular distance from centre to the line is equal to the radius of the circle.|– g – 2a| = $\dpi{100} \sqrt{g^{2} + f^{2} - c}$              ….(1)The circle cuts given circle orthogonally.$\dpi{80} \therefore$ 2g1g2 + 2f1f2 = c1 + c2$\dpi{80} \Rightarrow$ 2(-g)(0) + 2(-f)(0) = c – a2  $\dpi{80} \Rightarrow$  c = a2(1) becomes |g + 2a| = $\dpi{80} \sqrt{g^{2} + f^{2} - a^{2}}$Square both sides.g2 + 4ag + 4a2 = g2 + f2 – a2f2 = 4ag + 5a2Replace –g by x and –f by y  i.e.  g by –x and f by –y$\dpi{80} \therefore$ y2 = – 4ax + 5a2   i.e.y2 + 4ax – 5a2 = 0  is the required locus of the centres of the circles.

one year ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Analytical Geometry

View all Questions »

### 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