Circles...?

Question

We know that the standard equation of a circle is x^2+y^2=r^2, now my question is this-Is it possible to get the radius of the circle given the equation of the line from the origin P(0,0) to a certain point?How will you do this?

the p(0,0) is definitely inside the circle, if u graph the circle, the center is at (0,0) then a line drawn from the origin to any point is the radius, is it possible to get the radius given the equation of the line drawn from the origin to any point in the circle?

The measurement of the radius to be precise..

Answer 1

Let (a,b) be the given point, and it is on the circle. Then the distance of (a, b) from the origin is the radius of the circle.

The distance of (a, b) from the origin is sqrt[a^2 + b^2],
so r^2 = a^2 + b^2

Answer 2

what do you mean?

where the heck is the point? inside, outside, or on the circle?

In any case, is the circle of the form x^2 + y^2 = r^2

You would be better off giving the specifics. Vague questions can only lead to vague answers.
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I think you mean that a line is passing through the origin P (00) and a point Q in the interior of the circle.

Assuming the equation is standard form,

we find the points of intersection of the line with the circle by substituting x or y from the line equation into the equation of circle. We assumed the radius to be r.

So the points will come in terms of r, by solving the quadratic equations. Let us name them R and S

Now all you have to do is to use the slope formula, as

THE ORIGIN, THE POINT Q and the points R and S , whose co-ordinates you found above are in the same line .

So slope of line between R and Q equals the slope b/w Q and Origin,P. You get a quadratic and solve it.

I wish I could explain better, or could understand the question better,

Answer 3

Okay, in the case where the centre of the circle is not on the origin O(0,0), but some point (-f,-g), the equation of that circle would be given by x^2+y^2+2fx+2gy+c=0 and its radius would be given by sqrt[ (-g)^2+(-f)^2 -c ]. But here the centre is origin O(0,0). So its standard equation is x^2+y^2=r^2 and terms 2fx, 2gy, c are missing or all are equal to zero.
Lets assume that equation of line that joins centre O(0,0) and some point P(a,b)[ Assume some point P(3,4) ] on the circle is given to be bx-ay=0( or 4x-3y=0 ). Slope of this line is b/a (or 4/3) Slope of the line perpendicular to line bx-ay=0(4x-3y=0) and passing thro' P(a,b) would be given by ax+by-(a^2+b^2)=0(3x-4y-25=0). Now we have equation of the line passing thro' P(a,b) and a point O(0,0) outside the line. Apply formula, d(OP)= mod( [ ax1+by1-(a^2+b^2) ] / sqrt(a^2+b^2) ) where (x1,y1)=(0,0)
d(OP)=r= mod( sqrt(a^2+b^2) ) (r=mod( sqrt(3^2+4^2)=5)
Hence radius of such circle would be mod( sqrt(a^2+b^2) ).

Answer 4

we do? is that the new math?

Circle circumference is pi times radius squared