## samnatha Group Title 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

1. malevolence19

You need implicit differentiation $2x+\frac{dy}{dx}2y-2+2 \frac{dy}{dx}=0 \implies x-2+\frac{dy}{dx}(y+2)=0 \implies \frac{dy}{dx}=\frac{2-x}{y+2}$ $y-y_0=\left. \frac{dy}{dx} \right\|_{x_0,y_0}(x-x_0)$ So you need a point. Then you can find the equation then identify K

2. samnatha

see i can't do it that way as we have not done differentiation so it confuses me i know there is another way like you can rearrange the first equation so that x = .... and then sub it into the other one but after i do that i'm no sure what to do

3. malevolence19

If you just solve the first one for x: $x=-4y-K$ And plug it in you get: $(4y+K)^2+y^2+2(4y+K)+2y=0$ From this you can solve for K in terms of y. As long as you know the slope and everything is write for the original line this will give you the value which allows it to cross tangential. But the slope changes every small increment you move so I don't feel convinced that the slope (in general) can just be -1/4...

4. samnatha

how do i get the slope from that ?

5. samnatha

do i multiply out the equations so i get this $16y ^{2} + 8yk + k ^{2} + y ^{2} + 8y + 2k + 2y - 15$