equation of circle

👨🏻‍🏫 學術台數學

[匿名]

2013-04-22 2:53 am

1) A circle C1 passes through the intersection of the circle C: x^2 + y^2 = 9 and the line L: 2x-y+2=0. If P(1,-4) is on the circle C1, find the equation of C1.
Why is the equation of C1: x^2 + y^2 - 9 + k(2x-y+2)=0?

2) A tangent to the circle x^2+y^2=5 is drawn from the point (7,-2). What is the length of the tangent?

3)Consider the circle C: x^2+y^2-4x+2y-4=0 and the line L: x-y-6=0.
Find the equation of the circle which passes through the intersection of L and C and with the centre on the x-axis.

sol
let the required equation be x^2+y^2-4x+2y-4+k(x-y-6)=0

I don't understand.
Why do
{x^2+y^2-4x+2y-4+k(x-y-6)=0
{x-y-6=0
and
{x^2+y^2-4x+2y-4=0
{x-y-6=0
have the same points of intersection?

Please show your steps. Thanks.

更新1:

Appreciate it if you could show me the graph of Q2. Thanks.

更新2:

sorry, i don't get it "The reason is that C1 should fulfill x^2 + y^2 - 9=0; and 2x-y+2=0" what do you mean? how come x^2 + y^2 - 9 + k(2x-y+2)=0 ?

回答 (2)

Herman Tam

2013-04-22 9:53 am

✔ 最佳答案

Q1
Let the equation of C1: x^2 + y^2 - 9 + k(2x-y+2)=0
The reason is that C1 should fulfill
x^2 + y^2 - 9=0; and
2x-y+2=0
i.e. x^2 + y^2 - 9 + k(2x-y+2)=0
There are actually numerous possibilities, k can be 1, 2, 3, 4 or even 4.5, right? To find out the answer, hence the question should provide further information that (1, -4) is on C1.

Put (1, -4) into the equation:
1+16-9+k(2+4+2)=0
8+8k=0
k=-1
The equation of circle is x^2+y^x-2x+y-11=0

Question 2
The centre of the circle is (0,0)
The length of the point between (0,0) and (7, -2) is
(sqrt(49+4))=sqrt(53)
By pyth's theorem,
53=radius^2+ l^2
53=5+l^2
l^2=48
l=sqrt(48)

Question 3
Similar to Q1
Let the circle be x^2+y^2-4x+2y-4+K(x-y-6)=0
x^2+y^2+(k-4)x+(k+2)y-(6k+4)=0

Since the centre is on x-axis
the circle should be in form of
(x-a)^2+y^2=r^2
the coefficient of y should be equal to 0
i.e. k=-2
the equation of the circle is
x^2+y^2-6x+8=0

參考: ME

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[匿名]

2013-04-26 11:36 pm

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