Question

equation of circle centred at (1,2) and passing through (2,1) is​

Answer 1

Answer:

1.x^{2} +y^{2} +2gx+2fy+c=0 is the equation of circle and its centre is (-g,-f).

compare given centre (1,2) with (-g,-f) then

value of g is -1 and value of f is -2.

substitute in above equation we get

  2. x^{2} +y^{2} -2x-4y+c=0⇒1

substitute given point (2,1) in x and y places then we get

the value of c= -4-1+4+4=3

3.then substitute c value in equation 1

then the equation of circle is x^{2} +y^{2}-2x-4y+3=0

Answer 2

Answer:

(x-a)²+(y-b)²=r²

where (a,b) is the centre of the circle and r is the radius.

now,

We need to calculate the distance between these two points.

The formula for calculating the distance between two points is:

$d = \sqrt{(x2 - x1)^{2} + (y2 - y1) ^{2} }$

$d = \sqrt{(2 - 1)^{2} + (1 - 2)^{2} }$

$d = \sqrt{1 + 1}$

$d = \sqrt{2}$

now,

(x-1)²+(y-2)² = (√2)²

Therefore the equation of the circle is

(x-1)²+(y-2)²= 2

hope it helps yrr