r^1/2 is the radius of the circle.
r is the y-intercept of the line
y = r - x.
It is tangent to the circle if
r^2 = 2*(sqrtr)^2
(from Pythagorean rule)
Therefore r^2 = 2r
r = 2
Plot the two functions
y = 2-x and
x^2 + y^2 = 2
You will see that there is tangency at
(sqrt2, sqrt2)
For some positive real number r , the line x+y=r is tangent to the circle x^2+y^2 = r. Find r.
How do we do this? Set equal equations together??
4 answers
thanks a bunch sire
Yes set the two equations equal to each other. You then get that r = 2. I'm not going to tell you all of the work because drwls has already answered.
I am so confused by the symbols and square roots. I know what they mean, but there are so many of them!
Allow me to try to explain it without a surplus of symbols for people that don't frequent doing the math on the keyboard.
First, change the first equation to create a linear one. y=(-x)+r
The other equation is one of a circle. You should know this.
(x-h)²+(y-k)²=r²
Since x²+y²=r, the radius is √r.
That means the radius of the circle should touch the line at a point of tangency √r from the origin of the circle, 0,0. The radius which touches the point of tangency is perpendicular to y=(-x)+r.
Now you could do it the way drwls did, or you could use the formula for the length between a point and a line. Either way, you get r=2.
Allow me to try to explain it without a surplus of symbols for people that don't frequent doing the math on the keyboard.
First, change the first equation to create a linear one. y=(-x)+r
The other equation is one of a circle. You should know this.
(x-h)²+(y-k)²=r²
Since x²+y²=r, the radius is √r.
That means the radius of the circle should touch the line at a point of tangency √r from the origin of the circle, 0,0. The radius which touches the point of tangency is perpendicular to y=(-x)+r.
Now you could do it the way drwls did, or you could use the formula for the length between a point and a line. Either way, you get r=2.
1 answer