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The circle x^2+y^2-6x-10 y+k=0 does not ...

The circle `x^2+y^2-6x-10 y+k=0` does not touch or intersect the coordinate axes, and the point (1, 4) is inside the circle. Find the range of value of `kdot`

Text Solution

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The equation of the circle is
`x^(2)+y^(2)-6x-10y+k=0` (1)
whose center is C(3,5) and radius

`r=sqrt(34-k)`
If the circle does not touch or intersect the x-axis, then radius
`r lt y` coordinate of center C
or `sqrt(34-k)lt5`
or `34-klt25`
or `kgt9` (2)
Also, if the circle does not touch or intersect the y-axis, then the radius
`r lt x-` coordinate of center C
or `sqrt(34-k) lt3`
or `34 -k lt9`
or `kgt 25` (3)
If the point (1,4) is inside the circle, then
Its distance from center `C ltr`
or `sqrt((3-2)^(2)+(5-4)^(2))lt sqrt(34-k)`
or `5 lt34 -k`
or `klt29` (4)
Now, all the conditions (2),(3), and (4) are satisfied if `25 lt k lt29` which is the required range of the values of k.
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