# How do you find the distance from the point (2.1) to the circle #x^2 - 6x + y^2 + 2y= 0#?

##### 2 Answers

#### Answer:

#### Explanation:

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

#= (x-3)^2+(y-(-1))^2-10#

Adding

#(x-3)^2+(y-(-1))^2 = (sqrt(10))^2#

which is the standard form of the equation of a circle:

#(x-h)^2+(y-k)^2 = r^2#

with centre

The distance between the point

#sqrt((x_2-x_1)^2+(y_2-y_1)^2)#

#=sqrt((3-2)^2+(-1-1)^2) = sqrt(1+4) = sqrt(5) < sqrt(10)#

So the point

graph{(x^2-6x+y^2+2y)((x-2)^2+(y-1)^2-0.02)((x-3)^2+(y+1)^2-0.02) = 0 [-10, 10, -5, 5]}

#### Answer:

#### Explanation:

Given a circle

and a point

where

The

with

after simplifications we get at

or

or

Putting numeric values we get at

and after substitution in

and the nearest is