Question

11. Find the radius and centre of curvature of the curve x^2+y^2-2x-4y+4=0​

Answer 1

Answer:

Step-by-step explanation:

The standard eqn of a circle with center  ( a , b )  and radius  r is :

$( x - a )^2 + (y - b)^2 = r^2$

for  $x^2+y^2-2x-4y+4=0$

we need to complete the square.

$x^2 - 2x + y^2 - 4y + 4 = 0$

$(x^2 - 2x + 1^2) + ( y^2 - 4y + 2^2 ) - 1^1 - 2^2 + 4 = 0$

$(x - 1)^2 + (y + 2)^2 - 1 = 0$

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

$centre$ $=$ $(1 , -2)$

$radius =$ √ $1$

• if you don't mind i think the question is wrong somehow , if it's is right then my answer too is right but if your symbols or numbers are wrong then i don't know.

hope it helps you

Answer 2

Answer:

see the below plz

Step-by-step explanation:

The equation is

$x^2 + y^2 - 2x - 4y + 4 =0$

Complete the square

$x^2 - 2x + y^2 - 4y = 4$

$x^2 - 2x + 1 + y^2 - 4y + 4 = 4 + 1 + 4$

Factorize

$(x - 1)^2 + (y + 2)^2 = 9 = 3^2$

This is the equation of a circle, center  $C = (1 , 2 )$ and  radius $R = 3$

hope it helps you