Question

The centre and radius of the circle having equation
2x + 6y - x2 - y2 = 1 is
A) (1,3),3
B) (3, 1),3
C (1,3),4
D) (3, 1)

Answer 1

Answer:

A) (1,3),3

Step-by-step explanation:

$2x + 6y - x^2 - y^2 = 1\\or, x^2+y^2-2x-6y=-1\\or, x^2 - 2*x*1+1^2+y^2-2*y*3+3^2=-1+1^2+3^2\\or, (x-1)^2+(y-3)^2=3^2\\Compare with (x-h)^2+(y-k)^2=r^2\\Centre(h,k)=(1,3)\\Radius(r)=3 units$

Answer 2

Answer:

2x+6y−x

2

−y

2

=1

or,x

2

+y

2

−2x−6y=−1

or,x

2

−2∗x∗1+1

2

+y

2

−2∗y∗3+3

2

=−1+1

2

+3

2

or,(x−1)

2

+(y−3)

2

=3

2

Comparewith(x−h)

2

+(y−k)

2

=r

2

Centre(h,k)=(1,3)

Radius(r)=3units

Step-by-step explanation:

here is the answer