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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.A. a circleB. an ellipse which is not a circleC. a hyperbola0D. a parabola |
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Answer» Given equation ofcircle is `x^(2)+y^(2)-8x-8y-4=0`, whose centre is C(4, 4) and radius `=sqrt(4^(2)+4^(2)+4) = sqrt36 = 6` Let the centre of required circle be `C_(1)(x, y)`. Now, as it touch the X-axis, therefore its radius ` = |y|`. Also, it touch the circle `x^(2)+y^(2)-8x-8y-4=0 " therefore " "CC"_(1)=6+|y|` `rArr sqrt((x-4)^(2)+(y-4)^(2))=6+|y|` `rArr x^(2)+16-8x+y^(2)+16-8y=36+y^(2)+12|y|` `rArr x^(2)-8x-8y+32=36+12|y|` `rArr x^(2)-8x-8y-4=12|y|` Case I if `ygt0`, then we have `x^(2)-8x-8y-4=12y` `rArr x^(2)-8x-20y-4=0` `rArr (x-4)^(2)-20=20y` `rArr (x-4)^(2)=20(y+1)`, which is a parabola Case II if `ylt 0`, then we have `x^(2)-8x-8y-4=-12y` `rArr x^(2)-8x-8y-4+12y=0` `rArr x^(2)-8x+4y-4=0` `rArr x^(2)-8x-4=-4y` `rArr (x-4)^(2)=20-4y` `rArr(x-4)^(2)=-4(y-5)`, which is again a parabola . |
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