The straight line $x + y - 1 = 0$ meets the circle ${x^2} + {y^2} - 6x - 8y = 0$ at $A$ and $B$. Then, the equation of the circle of which $AB$ is a diameter is

$2\left( {{x^2} + {y^2}} \right) + 2y - 6 = 0$

$3\left( {{x^2} + {y^2}} \right) + 2y - 6 = 0$

JEE MATHEMATICS MCQ: SYSTEM OF CIRCLES

1 Easy

1 of 1

Application +4 | -1

The straight line

x + y - 1 = 0

meets the circle

{x^2} + {y^2} - 6x - 8y = 0

A

and

B

. Then, the equation of the circle of which

AB

is a diameter is

View answer and explanation

Correct option: A

{x^2} + {y^2} - 2y - 6 = 0

Option A (

{x^2} + {y^2} - 2y - 6 = 0

) is correct because it represents a circle with AB as a diameter, derived from the intersection of the given line and circle.

Practice All Chapter MCQs →

x = 1

{x^2} + {y^2} = 4

{x^2} + {y^2} + 8y - 4 = 0

{x^2} + {y^2} + gx + 4 = 0

x^2+y^2+3 x+2 y+1=0

x^2+y^2+3 x+4 y+2=0

{x^2} + {y^2} + 2ky + 2x + 6 = 0

{x^2} + {y^2} + 2ky + k = 0

k

x^2+y^2+2 a x+c y+a=0

x^2+y^2-3 a x+d y-1=0

P

Q

5 x+b y-a=0

P

Q

x + 2y - 3 = 0

{x^2} + {y^2} - 4y + 1 = 0

{x^2} + {y^2} - 4x - 2y + 4 = 0