Question

The ends of a quadrant of a circle have the
coordinates (1.3) and (3, 1) then the centre of the
such a citcle is
(A) (1.1)
(B) (2.2)
(C) (2.6)
(D) (4.4)​

Answer 1

Answer:

let A(1,3) and B(3,1) are points.

o(x,y) is centre.

AB^2=2^2+2^2

AB^2=8

OA^2+OB^2=8

But

OA=OB (radii)

so,

2OA^2=8

OA=2

So,radius=2

(x-1)^2+(y-3)^2=radius^2

x^2+1-2x+y^2+9-6y=4......(1)

since

both radii are equal

so,

(x-1)^2+(y-3)^2=(x-3)^2+(y-1)^2

x^2+1-2x+y^2+9-6y=x^2+9-6x+y^2+1-2y

so

-2x-6y=-6x-2y

4x=4y

x=y......(2)

from(1)&(2)

2y^2-8y+10=4

2y^2-8y+6=0

2y^2-6y-2y+6=0

(y-3)(2y-2)=0

y=3,1

so,

x=3,1

(3,3) is not in option

so(1,1) is answer

Answer 2

Plz refer to the attachment

hope it helps you

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