How do you write the equation for a circle with center at (0, 0) and touching the line 3x+4y=10?

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Answer 1 · 2016-06-13T00:15:49

#x^2 + y^2 =4#

To find the equation of a circle we should have the center and radius.
Equation of circle is:

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

Where (a,b) : are the coordinates of the center and
r :Is the radius

Given the center (0,0)

We should find the radius .
Radius is the perpendicular distance between (0,0) and the line 3x+4y=10

Applying the property of the distance #d# between line #Ax+By+C# and point #(m,n) # that says:

    # d = |A*m+B*n+C| / sqrt(A^2+B^2)#

The radius which is the distance from straight line #3x +4y -10=0 # to the center #(0,0) # we have:

A=3. B=4 and C=-10
So,

#r=#
#| 3*0 +4*0 -10| /sqrt( 3^2 + 4^2)#

= #| 0 +0-10| / sqrt(9 +16) #

= # 10/ sqrt(25)#
=#10/5#
=#2#

So the equation of the circle of center(0,0) and radius 2 is :

#(x-0)^2 + (y-0)^2 = 2^2#

That is # x^2 + y^2 =4#