Question

Equation of the circle having the lines by y^2-2y+4x-2xy=0 as its normal and passing through the point 2,1 is

Answer 1

Solution:

The normal line to circle is →y² - 2 y + 4 x -2 xy=0

→y(y-2) - 2 x(y-2)=0

→(y-2)(y-2 x)=0

the two lines are , y=2 and 2 x -y =0

The point of intersection of normals are centre of circle.

→ Put , y=2 in 2 x -y=0, we get

 →2 x -2=0

 →2 x=2

→ x=1

So, the point of intersection of normals is (1,2) which is the center of circle.

Also, the circle passes through (2,1).

Radius of circle is given by distance formula =$\sqrt{(1-2)^{2}+(2-1)^{2}} = \sqrt{1+1} =\sqrt{2}$

The equation of circle having center (1,2) and radius √2 is

    = (x-1)²+(y-2)²=[√2]²

→(x-1)²+(y-2)²= 2

Answer 2

Given:

Circle = y^2 -2y + 4x - 2xy = 0

Point ( 2, 1 )

To find:

The equation of the circle.

Solution:

Solving the circle's normal equation,

y ( y - 2 ) - 2x ( y - 2 ) = 0

( y - 2 ) ( y - 2x ) = 0

We get,

y = 2

2x - y = 0

Substituting,

2x - 2 = 0

2x = 2

x = 1

Hence, the point of intersection = ( 1, 2 )

To find the radius,

( ( x1 + x2 ) / 2, ( y1 + y2 ) / 2 ) )

Substituting,

We get,

Radius = √2

To find the equation of circle with center (1,2) and radius √2,

( x - 1 ^2 + ( y - 2 )^2 = [ √2 ]^2

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

Hence, the equation of the circle having the lines by y^2-2y+4x-2xy=0 as its normal and passing through the point 2,1 is ( x - 1 )^2 + ( y - 2 )^2 = 2