Explanation:
A Given that,
Equation of circle-
\(x^2+y^2-4 x-2 y-20=0 \text { and }\)
\(\text { Point }(10,7)\)
Now, \(\mathrm{x}=10, \mathrm{y}=7\) putting in above equation.
\(=100+49-4 \times 10-2 \times 7-20 .\)
\(=149-40-14-20=149-74=75\)
Here, we can see that the value is greater than 0 . the point \((10,7)\) lies outside the circle.
Now we can find the centre and radius of the circle.
\(x^2+y^2-2 \times 2 \times x-2 \times 1 \times y+2^2+1^2-4-1-20=0\)
\(\left(x^2-2 \times 2 \times x+2^2\right)+\left(y^2-2 \times 1 \times y+1^2\right)-4-1-20=0\)
\((x-2)^2+(y-1)^2=25\)
\((x-2)^2+(y-1)^2=25=5^2\)
\(\because \quad(x-h)^2+(y-k)^2=r^2\)
From equation (i) and (ii),
Centre ( \(\mathrm{h}, \mathrm{k}), \mathrm{r}=\) radius.
Here,
\(\mathrm{h}=2, \mathrm{k}=1, \quad(\mathrm{~h}, \mathrm{k})=(2,1)\)
Radius, \(\mathrm{r}=5\) unit.
The distance to the point \((10,7)\) from the center of the circle \((2,1)\) is given by
\(\mathrm{d}=\sqrt{\left(\mathrm{x}_1-\mathrm{x}_2\right)^2+\left(\mathrm{y}_1-\mathrm{y}_2\right)^2}\)
\(\mathrm{~d}=\sqrt{(10-2)^2+(7-1)^2}\)
\(\mathrm{~d}=\sqrt{64+36}=\sqrt{100}=10 \text { units. }\)
\(\because\) The least distance is given by -
\(l=\mathrm{d}-\mathrm{r}\)
\(l=10-5=5 \text { units }\)
The largest distance is -
\(l=\mathrm{d}+\mathrm{r}=10+5=15 \text { unit. }\)
