Home
Class 12
MATHS
The point of tangency of the circles x^2...

The point of tangency of the circles `x^2+ y^2 - 2x-4y = 0 and x^2 + y^2-8y -4 = 0`, is

Text Solution

Verified by Experts

The correct Answer is:
1
`C_(1)C_(2)=|r_(1)-r_(2)|`
Promotional Banner

Topper's Solved these Questions

  • JEE MAIN - 5

    VMC MODULES ENGLISH|Exercise PART III : MATHEMATICS (SECTION-2)|5 Videos

  • INVERSE TRIGONOMETRY

    VMC MODULES ENGLISH|Exercise JEE ADVANCED ( ARCHIVE )|10 Videos

  • JEE MAIN REVISION TEST - 30 | JEE -2020

    VMC MODULES ENGLISH|Exercise MATHEMATICS|25 Videos

Similar Questions

Explore conceptually related problems

The equation of the circle having its centre on the line x+2y-3=0 and passing through the points of intersection of the circles x^2+y^2-2x-4y+1=0a n dx^2+y^2-4x-2y+4=0 is

The common tangent at the point of contact of the two circles x^2+y^2-4x-4y=0, x^2+y^2+2x+2y=0 is

The common tangent at the point of contact of the two circles x^2+y^2-2x-4y-20=0, x^2+y^2+6x+2y-90=0 is

The equation of the circle having its centre on the line x+2y-3=0 and passing through the points of intersection of the circles x^2+y^2-2x-4y+1=0a n dx^2+y^2-4x-2y+4=0 is x^2+y^2-6x+7=0 x^2+y^2-3y+4=0 c. x^2+y^2-2x-2y+1=0 x^2+y^2+2x-4y+4=0

Find the equation of the circle passing through the point of intersection of the circles x^2 + y^2 - 6x + 2y + 4 = 0, x^2 + y^2 + 2x - 4y -6 = 0 and with its centre on the line y = x.

The equation of the circle passing through the point of intersection of the circles x^2+y^2-4x-2y=8 and x^2+y^2-2x-4y=8 and the point (-1,4) is (a) x^2+y^2+4x+4y-8=0 (b) x^2+y^2-3x+4y+8=0 (c) x^2+y^2+x+y=0 (d) x^2+y^2-3x-3y-8=0

The equation of the circle passing through the point of intersection of the circles x^2+y^2-4x-2y=8 and x^2+y^2-2x-4y=8 and the point (-1,4) is x^2+y^2+4x+4y-8=0 x^2+y^2-3x+4y+8=0 x^2+y^2+x+y=0 x^2+y^2-3x-3y-8=0

The equation of the circle passing through the point of intersection of the circles x^2+y^2-4x-2y=8 and x^2+y^2-2x-4y=8 and the point (-1,4) is x^2+y^2+4x+4y-8=0 x^2+y^2-3x+4y+8=0 x^2+y^2+x+y=0 x^2+y^2-3x-3y-8=0

The length of the common chord of the two circles x^2+y^2-4y=0 and x^2+y^2-8x-4y+11=0 is

The equation of the circle described on the common chord of the circles x^2 +y^2- 4x +5=0 and x^2 + y^2 + 8y + 7 = 0 as a diameter, is