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if the line 3x-4y-k=0 touches the circle `x^2+y^2-4x-8y-5=0` at (a,b),find (k+a+b)/5 is |
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Answer» Comparing the given equation of the circle with the general equation of the circle, `x^2+y^2+2gx+2fy+c = 0` We get, `g = -2 ` and `f = -4`. Any tangent that is touching the circle at `(x_1,y_1)` has the equation, `x x_1+yy_1+g(x+x_1)+f(y+y_1)+c = 0`Now, as the given line is touching the circle at `(a,b)` `:. ax+by-2(x+a)-4(y+b) -5 = 0` `=>(a-2)x+(b-4)y-(2a+4b+5) = 0` Comparing it with given equation, `3x-4y -k = 0`, `a-2 = 3, b-4 = -4 and k = 2a+4b+5` `a = 5, b = 0` `:. k = 10+0+5 = 15` So, `(k+a+b)/5 = (15+5+0)/5 = 20/5 = 4` |
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