Question 203604
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 {{{y = x+2}}}
Equation 2 is {{{(x-1)^2 + y^2 = 16}}}
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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 {{{y = x+2}}}
equation 2 says that {{{(x-1)^2 + y^2 = 16}}}
if you substitute (x+2) for y in equation 2, you get:
{{{(x-1)^2 + (x+2)^2 = 16}}}
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now you have one equation in one unknown that can be solved for x.
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since {{{(x-1)^2 = x^2 - 2*x + 1}}} and since {{{(x+2)^2 = x^2 + 4*x + 4}}} you substitute in equation:
{{{(x-1)^2 + (x+2)^2 = 16}}} to get:
{{{(x^2 - 2x + 1) + (x^2 + 4x + 4) = 16}}}
when you combine like terms on the left side of this equation, you get:
{{{2x^2 + 2x + 5 = 16}}}
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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:
{{{x^2 + x + 5/2 = 16/2}}}
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if you subtract 5/2 from both sides of this equation, you get:
{{{x^2 + x = 11/2}}}
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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 {{{(x+(1/2))^2}}}
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when you multiply this squaring factor out you get:
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{{{(x + (1/2))^2 = x^2 + x + (1/4)}}}
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since {{{(x + (1/2))^2 = x^2 + x + (1/4)}}} then:
{{{(x + (1/2))^2 - (1/4) =  x^2 + x}}}
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Your equation had already become:
{{{x^2 + x = (11/2)}}}
which can now become:
{{{(x+(1/2))^2 - (1/4) = (11/2)}}}
because {{{(x^2 + x)}}} and {{{((x+(1/2))^2 - (1/4))}}} are equivalent.
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your equation has now become:
{{{(x+(1/2))^2 - (1/4) = 11/2}}}
If you add (1/4) to both sides of this equation, you get:
{{{(x+(1/2))^2 = 11/2 + (1/4) = (23/4)}}}
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if you take the square root of both sides of this equation, you get:
{{{(x+(1/2))}}} = +/- {{{sqrt(23/4)}}}
if you subtract (1/2) from both sides of this equation, you get:
{{{x = -(1/2)}}} +/- {{{sqrt(23/4)}}}
which makes x =:
{{{-(1/2) + sqrt(23/4)}}}
or
{{{-(1/2) - sqrt(23/4)}}}
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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 = +/- {{{sqrt(15-x^2+2x)}}}
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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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{{{graph(600,600,-5,5,-4,4,sqrt(15-x^2+2*x),-sqrt(15-x^2+2*x),x+2)}}}
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