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samnatha

  • 3 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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  1. pansky
    • 3 years ago
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    diff the eqtn of the circle w.r.t. x to get dy/dx from circle = -(x-1)/(y+1) as gradient of line stated is -1/4 then put dy/dx = -1/4 and re-arrange to get y=4x-5. Where this line intersects circle the gradient=-1/4 . Then solve the eqtn of the circle and y=4x-5 as simultaneous etqns by substitution i.e replace all y values is eqtn of circle with 4x-5, giving x^2+(4x-5)^2-2x+2(4x-5)-15=0 expand and simplify to give x^2-2x=0 , factorise to get x(x-2)=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

  2. samnatha
    • 3 years ago
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    we haven't done differentiation yet so i don't know how to do it ?

  3. pansky
    • 3 years ago
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    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 (x-1)^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??

  4. samnatha
    • 3 years ago
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    no sorry it doesn't matter i'll get it eventually thanks for your help anyway tho :)

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