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Ebudo · Dec 25, 2009, 03:52 PM

please can you explain to me on how to solve this questions...
1, find the equation of the diameter of the circle x^2+y^2-8x+6y+21=0 which when produced passes through the point(2,1)...
2, find the equation of the circle which passes through the point (1,1)has a radius of 10 and whose centre lies on the line y=3x-7...

galactus · Dec 25, 2009, 04:47 PM

2, find the equation of the circle which passes through the point (1,1)has a radius of 10 and whose centre lies on the line y=3x-7...

To find the x-coordinate of the center, we can use the distance formula.

Since we are told that the center lies on y=3x-7, and the circle passes through $(x_{1},y_{1})=(1,1)$ , we have

$(y-y_{1})^{2}+(x-x_{1})^{2}=D^{2}$

The distance from (1,1) to the line at the center of the circle is then:

$((3x-7)-1)^{2}+(x-1)^{2}=100$

Solve for x, then sub that into the line equation to find y. This will be the coordinates of the center of the circle.

galactus · Dec 25, 2009, 06:12 PM

1, find the equation of the diameter of the circle x^2+y^2-8x+6y+21=0 which when produced passes through the point(2,1).

The find the equation of the circle from the given standard form, complete the square.

$(y-4)^{2}+(x+3)^{2}=4$

This has center (4,-3) and radius 2.

By 'equation of the diameter', I assume you mean the equation of the line that passes through the center of the circle, (4,-3) and
(2,1). I will leave you to find that. That is easy to do now.

Unknown008 · Dec 26, 2009, 04:17 AM

Small typo galactus :)

$(x-4)^2 + (y + 3)^2 = 4$

galactus · Dec 26, 2009, 08:24 AM

Thanks for the heads up, Unknown. I tried the reutation thing, but it wouldn't let me... as usual.

Ebudo · Dec 26, 2009, 01:37 PM

Thanks a lot.. galactus... its so easy d way u explained it... nd unknown, thnks 4 d correction... lol...