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# Equation to Circles

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Find the equation to the circle passing through (4,1) and (6,5).Centre of the circle is lying on 4x+y=16

Dec 28, 2014

#4
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Thanks Chris - It was a good question.  I liked it :)

Dec 29, 2014

#1
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Find the equation to the circle passing through (4,1) and (6,5).Centre of the circle is lying on 4x+y=16

Let (h,k) be the centre of the circle

So we know straight off that  4h+k=16      (1)

Now the distance from (h,k) to (4,1)  is equal to the distance of (h,k) to (6,5)

Use the distance formula (twice) to get an equation relating h to k

This will be a linear equatoin.    I will call it (2)

Now solve (1) and (2) simultaneously  to get the centre of the circle.

Now that you have the centre you can find the radius easily enough.

and once you have the centre and the radius you just plug the values into the formula to the general circle equation to get your specific circle equation.

Can you take it from there?

I don't mind helping more but learning should be an interactive process so I want input from you first. Dec 28, 2014
#2
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Nice job, Melody......I couldn't think how to get this one started...!!!!...I might work through it myself, just to see if I can do it....!!!   Dec 29, 2014
#3
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Using 4h + k = 16....we have that k = 16 - 4h.....and equating distances, we have

(h-4)^2 +  ( 16 - 4h -1)^2  = (h-6)^2 + (16 - 4h - 5)^2 ....... simplify

(h-4)^2 + (15 - 4h)^2  = (h-6)^2 + (11-4h)^2

h^2 - 8h + 16 + 16h^2 - 120h + 225 = h^2 - 12h + 36 + 16h^2 - 88h + 121

-8h + 16 -120h + 225  =  -12h + 36 -88h + 121

-128h + 241 = -100h + 157

84 = 28h

h = 3

So...using k = 16 - 4h....then k = 16 - 4(3)  = 16 - 12 = 4

Then (h,k)  = (3, 4)  and the circle has a radius of  √10

Here's a pic......https://www.desmos.com/calculator/ablcil3bms   Dec 29, 2014
#4
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