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The centres of those circles which touch the circle, `x^2+y^2-8x-8y-4=0`, externally and also touch thex-axis, lieon :(1) a circle.(2) an ellipse which is not acircle.(3) a hyperbola.(4) a parabola. |
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Answer» `x^2 + y^2 - 8x - 8y - 4= 0` `x^2 - 8x + 16 + y^2 - 8y+16 - 4 - 16-16= 0` `(x-4)^2 + (y- 4)^2 - 36 = 0` `(x-4)^2 + (y-4)^2 = 6^2` radius`= 6` centre= `(4,4)` centre of circle which touches the given circle externally is `sqrt((h-4)^2 + (k-4)^2) = 6 + r` `sqrt((h-4)^2 + (k-4)^2) = (6+k)` `(h-4)^2 + (k-4)^2 = (6 + k)^2` `h^2 + 16 - 8h + k^2 + 16 - 8k = 36 + k^2 + 12k` `h^2 -8h-20k + 32 - 36= 0` `h^2 - 8h - 20k - 4= 0` keeping h= x & k=y `x^2 - 8x - 20y - 4=0` this eqn is of parabola so, option 3 is correct |
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