Circles!

hellllo!!!
i neeeeeeeeed some help plz!

A circle C1 has equation
x^2 + y^2 – 12x + 4y + 20 = 0.

The circle C1 cuts the x-axis at the points A and B.

Find an equation of the circle C2 with diameter AB.

Thanks...

Reply 1

cms271828

Hi, no worries, its' easy.

On C1, let y=0.
This gives x^2-12x+20=0
then (x-2)(x-10)=0
so A=(2,0) and B=(10,0) or vice-versa.

Since AB is a diameter of C2, then radius of C2=
(10-2)/2=4.
Centre of C2=((2+10)/2,(0+0)/2)=(6,0)

Hence using (x-x1)^2+(y-y1)^2=r^2. gives the equation of C2 as:
(x-6)^2+(y-0)^2=4^2
hence x^2-12x+y^2+20=0.

OK?

Reply 2

Font

cms271828

(x-6)^2+(y-0)^2=4^2
hence x^2-12x+y^2+20=0.

OK?

the thing above looks right...but i don't know why you said hence the orig. equation?

Pk

Reply 3

yazan_l

x²+y²-12x+4y = -20
x²-12x+36 + y²+4y+4 = 20
(x-6)² + (y+2)² = 20
radius of C1 = sqrt[20]
centre of C1 = (6,-2)

centre of C2 is on the x-axis ==> y = 0
and it's above the centre of C1 ==> x=6

centr of C2 = (6,0)

AB = 2 . sqrt[20-2²] = 2.sqrt[16] = 8
==> radius of C2 = 1/2 AB = 4
==> eq. of C2 is : (x-6)² + y² = 16

Reply 4

ossoss87

thanks alot!!!!!!!!

Reply 5

cms271828

Phil23

the thing above looks right...but i don't know why you said hence the orig. equation?

Pk

Its not,there is a 4y in original equation

Reply 6

yazan_l

cms271828

Its not,there is a 4y in original equation

well it can be , and easier, written in the form (x-a)²+(y-b)² = r²
so there is no need to go any further! they give full mark for such answers unless they said to give it in the other form

Reply 7

cms271828

yazan_l

well it can be , and easier, written in the form (x-a)²+(y-b)² = r²
so there is no need to go any further! they give full mark for such answers unless they said to give it in the other form