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Let p and q be constants such that the graph of x^2 + y^2 - 6x +py + q is tangent to the y-axis. What is the area of the region enclosed by the graph?

 

So I have moved it around to be the standard equation to finding the radius of a circle (x-h)^2+(y-k)^2=r^2
And i ended up with (x-3)^2+(y+(p/2))^2=-q-9-(p/2)^2
I have the feeling that I'm over complicating this could anyone give me some help?

 May 17, 2019
 #1
avatar+8829 
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I think you’re pretty much on the right track smiley

 

Assuming:   \(x^2 + y^2 - 6x + py + q = 0\)

 

\(x^2 + y^2 - 6x + py + q = 0\)
                                                    Subtract  q  from both sides of the equation
\(x^2 + y^2 - 6x + py = -q\)
                                                    Rearrange the terms on the left side of the equation.
\(x^2 - 6x + y^2 + py = -q\)
                                                                                          Add  9  and add  \((\frac{p}{2})^2\)  to both sides of the equation.

\(x^2 - 6x + 9 + y^2 + py + (\frac{p}{2})^2 = -q + 9 + (\frac{p}{2})^2\)
                                                                                           Factor both perfect square trinomials on the left side.
\((x - 3)^2 + (y + \frac{p}{2})^2 = -q + 9 + (\frac{p}{2})^2\)

 

Now we can see that the center of the circle is the point  (3, -\(\frac{p}{2}\))

 

Because the circle is tangent to the y-axis,

 

radius  =  distance between center and y-axis

radius  =  distance between  (3, -\(\frac{p}{2}\))  and  (0, -\(\frac{p}{2}\))

radius  =  3

 

area  =  π · radius2

area  =  π · 32

area  =  9π     sq. units

 

You can play around with the sliders on this graph to check:

https://www.desmos.com/calculator/zvm1f8xhaf

 

Notice that the distance between the center and the y-axis stays  3 .

 May 18, 2019

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