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samnatha
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find the value of K for which x + 4y + K = 0 is a tangent to x^2 + y^2  2x + 2y  15 = 0
 one year ago
 one year ago
samnatha Group Title
find the value of K for which x + 4y + K = 0 is a tangent to x^2 + y^2  2x + 2y  15 = 0
 one year ago
 one year ago

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campbell_st Group TitleBest ResponseYou've already chosen the best response.0
you need to find the point(s) of contact before you can find K you could solve simultaneously and finding the 1 point of contact between the curve and the tangent. Substitution seems the most obvious method. An alternative method is to differentiate and let the derivative equal the slope of the tangent (1/4) then solve for x.. once you know x find y by substituting. When you have the point of contact substitute into the tangent equation and solve for K
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.0
i tried your first method and subed in 4y  k for x i then got a big long equation that went like this 17y^2 + K^2 + 8yk + 2K + 10 y = 0
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
ok because you are dealing with a circle then there are 2 points where the the tangents can occur dw:1354733396218:dw so rewrite theequation as 2 parts \[y = \sqrt{17  (x 1)^2}  1... and ....y =  \sqrt{17  (x1)^2} 1\] differentiate both of those.... let the derivatives equal 1/4 and solve for x.
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.0
we have not done differentiate yet so i have no idea how to do it that way
 one year ago

campbell_st Group TitleBest ResponseYou've already chosen the best response.0
ok.... that makes it difficult..
 one year ago

samnatha Group TitleBest ResponseYou've already chosen the best response.0
yeah it does how would i do it the other way do u know ?
 one year ago
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