Home
Class 12
MATHS
Statement-1: The equation of a circle th...

Statement-1: The equation of a circle through the origin and belonging to the coaxial system, of which limiting points are (1, 1) and (3, 3) is `2x^(2)+2y^(2)-3x-3y=0`
Statement-2: The equation of a circle passing through the points (1, 1) and (3, 3) is `2x^(2)+y^(2)-2x-6y+6=0`.

A

Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.

B

Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.

C

Statement-1 is True, Statement-2 is False.

D

Statement-1 is False, Statement-2 is True.

Text Solution

Verified by Experts

The correct Answer is:
B
Statement-1 is true
Statement-2 is also true as the circle in it passes through(1, 1) and (3, 3) but does not lead to statement-1.
Promotional Banner

Topper's Solved these Questions

  • CIRCLES

    OBJECTIVE RD SHARMA|Exercise Exercise|132 Videos

  • CIRCLES

    OBJECTIVE RD SHARMA|Exercise Chapter Test|55 Videos

  • CIRCLES

    OBJECTIVE RD SHARMA|Exercise Section-I (Solved MCQs)|1 Videos

  • CARTESIAN PRODUCT OF SETS AND RELATIONS

    OBJECTIVE RD SHARMA|Exercise Chapter Test|31 Videos

  • COMPLEX NUMBERS

    OBJECTIVE RD SHARMA|Exercise Chapter Test|59 Videos

Similar Questions

Explore conceptually related problems

Find the equation of a circle which passes through the point (3, 2) and concentric with the circle x^(2)+y^(2)-4x +2y - 1=0 .

Find the equation of the circle passing through the point (-2,1) and touching the line 3x-2y-6=0 at the point (4,3).

The equation of circle passing through the point (1, 1) and point of intersection x^(2) + y^(2) = 6 and x^(2) + y^(2) -6x + 8 =0 , is

Find the equation of the circle which passes through the points (-1,2) and (3,-2) and whose centre lies on the line x-2y=0

Find the equation of the circle passing through the point (1,9), and touching the line 3x + 4y + 6 = 0 at the point (-2,0) .

Equation of circle touching x+y-2=0 at (1,1) and passing through (2,-3) is

An equation of the normal at the point (2,3)to the circle x ^(2)+ y^(2) - 2x - 2y - 3=0 is

The vector equation of the line passing through the point (-1, -1 ,2) and parallel to the line 2x - 2= 3y +1 = 6z -2 is

The equation of the plane passing through the points (0,1,2) and (-1,0,3) and perpendicular to the plane 2x+3y+z=5 is