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Question 203604: Please someone help me about this :
Find the point of the intersection of the line y=x+2 and the circle (x-1)^2 + y^2=16
I guess i should find a point on the line which has distance 4 unit from the centre of the circle that is (1,o) but i don't know how!
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the point of the intersection of the line y=x+2 and the circle
(x-1)^2 + y^2=16
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Substitute for "y" to get:
(x-1)^2 + (x+2)^2 = 16
x^2-2x+1 + x^2+4x+4 = 16
2x^2 +2x + 5 = 16
2x^2 + 2x - 9 = 0
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x = [-2 +- sqrt(4 - 4*2*-9)]/4
x = [-2 +- sqrt(4 + 72)]/4
x = [-2 +- 2sqrt(19)]/4
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x = (-1/2) + (1/2)sqrt(19) or x = (-1/2)-(1/2)sqrt(19)
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Substitute each of these into y = x+2 to get the y-values corresponding to the
x values:
Those y values will be (3/2)+(1/2)sqrt(19) and (3/2)-(1/2)sqrt(19)
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Cheers,
Stan H.
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Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! Your problem:
Find the point of the intersection of the line y=x+2 and the circle (x-1)^2 + y^2=16
I guess i should find a point on the line which has distance 4 unit from the centre of the circle that is (1,o) but i don't know how!
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It looks like you have 2 equations to deal with.
Equation 1 is 
Equation 2 is 
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In order to find the points of intersection, you need to find the values of x and y that satisfy both equations simultaneously.
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A relatively simple way to do this is to take the value of y in equation 1 and substitute it for the value of y in equation 2.
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equation 1 says that
equation 2 says that
if you substitute (x+2) for y in equation 2, you get:
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now you have one equation in one unknown that can be solved for x.
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since  and since  you substitute in equation:
to get:
when you combine like terms on the left side of this equation, you get:
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you can solve this by completing the square.
You could also solve this by using the quadratic formula.
We'll be solving it by completing the square.
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if you divide both sides of this equation by 2, you get:
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if you subtract 5/2 from both sides of this equation, you get:
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take ½ the b term which is the coefficient of x which is 1 and you get ½.
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Your squaring factor on the left side of the equation will be 
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when you multiply this squaring factor out you get:
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since then:
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Your equation had already become:
which can now become:
because  and  are equivalent.
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your equation has now become:
If you add (1/4) to both sides of this equation, you get:
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if you take the square root of both sides of this equation, you get:
 = +/- 
if you subtract (1/2) from both sides of this equation, you get:
 +/-
which makes x =:
or
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This becomes:
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x = -2.8979
or
x = +1.8979
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We can use these values of x to solve for y.
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The equation of:
(x-1)^2 + y^2 = 16
becomes:
y = +/- 
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This equation can also be used to graph the problem which will be shown below after all the algebra has been taken care of.
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if x = -2.8979, then y = +/- .897915762
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if x = 1.8979 then y = +/- 3.8979
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Of these possible values for y:
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y = -0.8979 satisfies both equations when x = -2.8979
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y = + 3.8979 satisfied both equations when x = +1.8979
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These look like your points of intersection of the line with the circle.
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The intersection points are represented by the ordered pair (x,y)
One point of intersection is (-2.8979,-0.8979)
The other point of intersection is (1.8979,3.8979)
A graph of the equation of the circle and a graph of the equation of the line are shown below and confirm the algebra.
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