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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
 one year ago
 one year 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
 one year ago
 one year ago

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panskyBest ResponseYou've already chosen the best response.1
diff 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
 one year ago

samnathaBest ResponseYou've already chosen the best response.0
we haven't done differentiation yet so i don't know how to do it ?
 one year ago

panskyBest ResponseYou've already chosen the best response.1
wow 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??
 one year ago

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