What Is the Equation of a Circle Tangent to Both Axes in the 4th Quadrant?

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To find the equation of a circle tangent to both axes in the fourth quadrant and passing through the point (8, -9), the center of the circle must be at (a, b) where the radius r equals both a and b. The general equation of the circle simplifies to (x-a)² + (y-a)² = r². By substituting the coordinates (8, -9) into this equation, a quadratic equation for a can be derived. This results in potentially two valid solutions for the radius and center of the circle. The key takeaway is that the radius must equal the x and y coordinates of the center for the circle to be tangent to both axes.
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1. Find the equation of a circle tangent to both axis with its center in the 4th quadrant passing through (8,-9)



2.Equation of Circle (X-a)^2+(Y-b)^2)=r^2



3. I tried solving graphically, but am not sure whether my graph is precise

Thanks for the help
 
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Think about the fact that the circle is "tangent to both axis ". If a circle has center at (a, b) and is tangent to the y-axis, then the line from (a, b) to (0, b) (the point of tangency) is a radius- and has length a. If a circle has center at (a, b) and is tangent to the y-axis, then its radius is a.

If a circle has center at (a,b) and is tangent to the x-axis, then the line from (a, b) to (a, 0) (the point of tangency) is a radius- and has length b. If a circle has center at (a, b) and is tangent to the x-axis, then its radius is b.

Now, what if a circle has center at (a, b) and is tangent to both axes? Then you get that the radius is r= a= b. That is, you know that the general equation of a circle, (x-a)2+ (y- b)2= r2, for this case becomes (x-a)2+ (y- a)2= r2 so you only need to find a.

You are told that the circle passes through (8, -9) so you know that x= 8, y= -9 satisfies that equation. Put x= 8, y= -9 and solve for a. You will get a quadratic equation for a so there may be 2 correct answers.
 

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