Question

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

Answer 1

`x^2 + y^2 =4`

Explanation:

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`