What is the equation of a circle that has its center at the origin and is tangent to the line with equation 3x-4y=10?

1 Answer
Aviv S.
May 5, 2018

The circle's equation would be #x^2+y^2=4#

Explanation:

We know that a circle's equation centered at the origin would look something like #x^2+y^2=r^2#, where #r# is the radius. We want to find #r# so that we can have the circle's full equation.

We can make a system of equations:

#{(3x-4y=10),(x^2+y^2=r^2):}#

We can solve for #y# in the first equation:

#3x-4y=10#

#-4y=10-3x#

#4y=3x-10#

#y=(3x-10)/4#

We can put this into the second equation:

#x^2+((3x+10)/4)^2=r^2#

#x^2+(3x+10)^2/4^2=r^2#

#x^2+(9x^2+60x+100)/16=r^2#

#(16x^2)/16+(9x^2+60x+100)/16=r^2#

#(25x^2+60x+100)/16=r^2#

#25/16x^2+60/16x+100/16=r^2#

#25/16x^2+15/4x+25/4=r^2#

Now we have to think a little. The word "tangent" in the question would mean that the system of equations has one solution for #x# and #y#. Therefore, we want this above quadratic to have one solution.

In order for a quadratic to have exactly one solution (and not two or zero), the discriminant has to be #0#. We can identify the #a#, #b#, and #c# values of the quadratic, and set the discriminant (#b^2-4ac#) equal to #0#.

#underbrace(25/16)_ax^2+underbrace(15/4)_bx+underbrace(25/4-r^2)_c=0#

Setting the discriminant equal to #0#:

#(15/4)^2-4(25/16)(25/4-r^2)=0#

#225/16-25/4(25/4-r^2)=0#

#225/16-625/16+25/4r^2=0#

#-400/16+25/4r^2=0#

#25/4r^2=400/16#

#25r^2=1600/16#

#25r^2=100#

#r^2=4#

#r=2#

We found the radius! We can now write the equation of the circle:

#x^2+y^2=2^2#

#x^2+y^2=4#

That's the solution. Hope this helped!

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