how do you find the equation of a circle w/ radius =5 and tangent to the positive x-and y-axes?

Answer 1

If the x and y axes are tangents to the circle then the radius is perpendicular to the axes at that point. [Draw a quick sketch]

Since the radius is 5, the centre of the circle is at (5,5)

Equation will be (x - 5)^2 + (y - 5)^2 = 5^2

Answer 2

so as to locate the equation of a line, you choose the slope and a factor which you comprehend is on the line. The slope is user-friendly: a tangent to a circle is perpendicular to the radius on the factor the place the line would be tangent to the circle. In different words, the radius of your circle starts at (0,0) and is going to (3,4). This slope is comparable to upward thrust/run = (y2-y1)/(x2-x1) = (4-0)/(3-0) = 4/3. A line this is perpendicular to this radius line could have a slope this is the unfavourable reciprocal of the slope you comprehend. In different words, the line we choose could have a slope of -3/4. the factor all of us comprehend is on the line is (3,4). So the established equation of the line is: (y-y0) = m(x-x0), the place (x0, y0) is the factor all of us comprehend is on the line, or (3,4) to that end m is the slope, -3/4 to that end x,y stay as variables (y - 4) = (-3/4)(x - 3) y - 4 = (-3/4)x + 9/4 y = (-3/4)x + 25/4

Answer 3

This is a circle with the center at (5,5) and radius 5, thus
using the eqn of a circle

(x - h)^2 + (y-k)^2 = r^2

where h and k are the x- and y-coordinates of the center of the circle and r is the radius

(x - 5)^2 + (y - 5)^2 = 25

Answer 4

Centre (5 , 5)
r = 5
(x - 5)² + (y - 5)² = 25 is equation of this circle.