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

2 Answers
George C.
May 30, 2016

#sqrt(10)-sqrt(5) ~~ 0.9262#

Explanation:

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

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

Adding #10# to both sides and transposing, we can express this as:

#(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 #(h, k) = (3, -1)# and radius #r = sqrt(10)#

The distance between the point #(2, 1)# and the centre of the circle #(3, -1)# is given by the distance formula:

#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 #(2, 1)# is inside the circle at a distance of #sqrt(10)-sqrt(5)# from the circumference.

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]}

Cesareo R.
May 30, 2016

#d=0.92621#

Explanation:

Given a circle

#C->(x-x_c)^2+(y-y_c)^2 = r^2#

and a point #q_0=(x_0,y_0)#, the minimum distance from #q_0# to the circle #C# is given by

#d = norm(q_0-p_0)#

where #p_0# is the intersection of #C# and the straight

#S->s = q_0+lambda(p_c-q_0)#

The #C# equation can be written as #C->(p-p_c).(p-p_c) = r^2#

with #p = (x,y)#. In the intersection occurs #p = s# then substituting

#C nn S ->(q_0+lambda(p_c-q_0)).(q_0+lambda(p_c-q_0))=r^2#

after simplifications we get at

#lambda^2-2lambda+1=r^2/((p_c-q_0).(p_c-q_0))#

or

#(lambda-1)^2=r^2/((p_c-q_0).(p_c-q_0))#

or

#lambda = 1pmr/sqrt((p_c-q_0).(p_c-q_0))#

Putting numeric values we get at

#lambda = 1 pm sqrt(2)#

and after substitution in #S#

#p_0^1=(2 sqrt[2] + 3 (1 - sqrt[2]), -1 + 2 sqrt[2])#
#p_0^2 = (-2 sqrt[2] + 3 (1 + sqrt[2]), -1 - 2 sqrt[2])#

and the nearest is #q_0^1# giving a distance of

#norm(p_0^1-q_0)=0.92621#

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