Question

find the equation of a circle that is tangent to the x-axis at (5,0) and tangent to the y-axis at (0,-5).

Answer 1

A cricle that is tangent to the two axes at the given points (5,0) and (0,-5) will line in the 4th quadrant of the plane, with it's center at (5,-5) and a radius of 5 units (diameter 10 units.) With this informatino we can plug in know value to the circle equation to get out solution.

The circle equation is: 
$(x-h)^{2} +(y-k)^{2} = r^{2}$ 
where 'r' is the radius and (h,k) is the circle's center, Thus:

$(x-5)^{2} +(y-(-5))^{2} = 5^{2}$ 

This can be simplified to:
$(x-5)^{2} +(y+5)^{2} =25$  

Answer 2

The equation should be $(x - 5)^2 + (y + 5)^2 = 25$

Given that,

• tangent to the x-axis at (5,0) and tangent to the y-axis at (0,-5).

Based on the above information, the equation is as follows:

The circle equation is:

$(x - h)^2 + (y - k)^2 = r^2\\\\ (x - 5)^2 + (y - (-5))^2 = 5^2\\\\ (x - 5)^2 + (y + 5)^2 = 25$

Here r be the radius and (h,k) is the circle's center

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