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
May 5, 2018

The circle's equation would be x^2+y^2=4x2+y2=4

Explanation:

We know that a circle's equation centered at the origin would look something like x^2+y^2=r^2x2+y2=r2, where rr is the radius. We want to find rr 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!