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...

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.

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.

Small typo galactus :)

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

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

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