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samnatha
 2 years ago
pleasssse helpp me !
find the value of K for which x + 4y + K = 0 is a tangent to x^2 + y^2  2x + 2y  15 = 0
samnatha
 2 years ago
pleasssse helpp me ! find the value of K for which x + 4y + K = 0 is a tangent to x^2 + y^2  2x + 2y  15 = 0

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pansky
 2 years ago
Best ResponseYou've already chosen the best response.1diff the eqtn of the circle w.r.t. x to get dy/dx from circle = (x1)/(y+1) as gradient of line stated is 1/4 then put dy/dx = 1/4 and rearrange to get y=4x5. Where this line intersects circle the gradient=1/4 . Then solve the eqtn of the circle and y=4x5 as simultaneous etqns by substitution i.e replace all y values is eqtn of circle with 4x5, giving x^2+(4x5)^22x+2(4x5)15=0 expand and simplify to give x^22x=0 , factorise to get x(x2)=0 , therefore x=0 or x=2, hence tangent with gradient of 1/4 happens at x=0 or x=2 and putting these values back into circle gives y values of 5 and 3 respectively. therefore only two possible coordinates are (0,5) and (2,3) . Put these back into eqtn of line x+4y+k= 0 and you get k=20 or k=14. These solutions can be verified with any graphing calculator or programme. Please note there are two solutions, not one, unless specified that k>0

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.0we haven't done differentiation yet so i don't know how to do it ?

pansky
 2 years ago
Best ResponseYou've already chosen the best response.1wow not sure how they expect you to do it then..... you have two simultaneous equations with 3 variables.....you can't solve that. .... you could draw it, but circle is (x1)^2 + (y+1)^2 17 =0 so its a circle centre (1,1) radius square root of 17, so you can't draw that exactly....... the only other way is trial and error with two equations and adjust k until you find solution but that's a waste of your time with no mathematical insight gained....so no other information given at all??

samnatha
 2 years ago
Best ResponseYou've already chosen the best response.0no sorry it doesn't matter i'll get it eventually thanks for your help anyway tho :)
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