The $$x^{2} + y^{2}- 6x - 10y + k = 0$$ does not touch or intersect the coordinate axes. If the point (1, 4) does not lie outside the circle, and the range of k is (a, b] then a + b is

The given equation of circle is $$x^{2} + y^{2}- 6x - 10y + k = 0$$

On comparing with the standard equation, $$x^{2} + y^{2}+ 2gx + 2fy + c = 0$$

We get $$g=-3,\ f=-5$$ and $$c=k$$

Centre of circle is $$\left(-g,-f\right)=\left(3,5\right)$$

Radius of circle is $$\sqrt{\ g^2+f^2-c}=\sqrt{\ 3^2+5^2-k}=\sqrt{\ 34-k}$$

Now, it is given that the circle does not touch or intersect the coordinate axes.

So, we can say, Distance of the centre from both the coordinate axes will be greater than radius.

Now centre is $$\left(3,5\right)$$

So, distance of the centre from x-axis=3 units and from y-axis=5 units

So, $$3>\sqrt{\ 34-k}$$ and $$5>\sqrt{\ 34-k}$$

So, $$k>25$$ and $$k>9$$

So, overall $$k>25$$ ------>(1)

Also, (1, 4) does not lie outside the circle

This means the point (1,4) can lie either on the circle or inside the circle.

So, we can say distance of the point (1,4) from the centre will be lesser than or equal to radius.

Now, distance of point (1,4) from centre (3,5) is $$\sqrt{\ \left(1-3\right)^2+\left(4-5\right)^2}=\sqrt{\ 5}$$

So, $$\sqrt{\ 5}\le\sqrt{\ 34-k}$$

or, $$k\le29$$ ----->(2)

From (1) and (2), we can say, $$k$$ lies in the interval $$\left(25,29\right]$$

So, $$a=25$$ and $$b=29$$

So, $$a+b=25+29=54$$