The centres of those circles which touch the circle, `x^2+y^2-8x-8y-4=0` , externally and also touch the x-axis, lie on : (1) a circle. (2) an ellipse which is not a circle. (3) a hyperbola. (4) a parabola.

Options

A. a circle
B. an ellipse which is not a circle
C. a hyperbola0
D. a parabola

Explanation

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 .