Circles

[Lyphta]

Determine an equation of the form x^2+y^2+ax+by+c=0 for the circle with center C and passing through point P.

C(3,-2)
P(7,1)

 

[galactus]

Use \(\displaystyle r=\sqrt{(7-3)^{2}+(1+2)^{2}}]\)

Now use the standard equation of a circle with h=3 , k=-2 , r=the above.

By squaring terms and simplifying you can write it in the form you require.

 

[Lyphta]

thanks galactus!!

if i could ask one more question just not on circles but on parabolas real fast. i was wondering how you determine the equation of the parabola with the vertex at the origin and the focus at (10,0)

 

[Unco]

The standard form of a right horizontal parabola is
\(\displaystyle (y - y_0)^2 = 4p(x - x_0)\)
where it is centred on \(\displaystyle (x_0, \, y_0)\) and has a focal distance of \(\displaystyle p\).

 

[Lyphta]

so you mean that i plug in the origin (0,0) and the focus of (10,0) into the (y-y)=4p(x-x)? cause i got... x=5/2y^2 when the answer is y^2=40x...

 

[Unco]

The focal length, \(\displaystyle p\), is 10.

 

[Lyphta]

thanks Unco! i just have a question as to how you got that equation or formula.

oh and if we are solving for the directrix, is it the same formula?

 

[Unco]

That is the standard equation of a right horizontal parabola.

An upright vertical parabola has the equation \(\displaystyle (x - x_0)^2 = 4p(y - y_0)\)

If you are dealing with the focus, directrix et al this is the form you want to use.

Here, the directrix is a vertical line one focal length, 10, 'behind' the parabola. So it has equation x = -10.

 

[Lyphta]

oh i see! thanks Unco. cause my teacher gave me no real formula equation thing. thanks so much