MHB Equation of the Circle (Part 2)

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SUMMARY

The equation of the circle tangent to the y-axis with center at (3, 5) is derived using the standard circle equation \((x-h)^2+(y-k)^2=r^2\). Given that the radius \(r\) equals the absolute value of the x-coordinate of the center, we find \(r=|3|=3\). Substituting the values into the equation yields \((x-3)^2+(y-5)^2=3^2\), which simplifies to \((x-3)^2+(y-5)^2=9\). This represents the required equation of the circle.

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  • Explore the concept of tangents to circles in coordinate geometry
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Find the equation of the circle tangent to the y-axis and with center (3, 5).

Can someone provide the steps needed to solve this problem?
 
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The equation of a circle centered ar $(h,k)$ is given by:

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

If the circle is tangent to the $y$-axis, then its radius must be $r=|h|\implies r^2=h^2$, thus we have:

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

We are given $(h,k)=(3,5)$, so plug in those numbers. :D
 
MarkFL said:
The equation of a circle centered ar $(h,k)$ is given by:

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

If the circle is tangent to the $y$-axis, then its radius must be $r=|h|\implies r^2=h^2$, thus we have:

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

We are given $(h,k)=(3,5)$, so plug in those numbers. :D

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

(x - 3)^2 + (y - 5)^2 = 3^2

(x - 3)^2 + (y - 5)^2 = 9
 

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