0

Guest

Courses New

The equation of circle with centre (4,3) and touching x^2+y^2=1 is

The equation of circle with centre (4,3) and touching x^2+y^2=1 is

Grade:12th pass

1 Answers

venkateshwarrao
17 Points
7 years ago
C1=(0,0)&r1=1,C2=(4,3) since the two circles touch each other externally then c1c2=r1+r2Implies 5=1+r2 implies r2=4Therefore the equation of the circle is(X-4)^2+(y-3)^2=16x^2+y^2-8x-6y+9=0

Think You Can Provide A Better Answer ?

Other Related Questions on Analytical Geometry

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More