The position of the point (1, 2) with respect to the circle

​$$x^2+y^2-3x-4y+1=0 :$$​​

Correct option is D

Given:

The equation of the circle is:

​$$x^2 + y^2 - 3x - 4y + 1 = 0$$​​

We need to determine the position of the point (1, 2) with respect to this circle.

Concept Used:

If result = 0 → Point lies on the circle

If result < 0 → Point lies inside the circle

If result > 0 → Point lies outside the circle

Solution:
Substitute x = 1, y = 2 into the equation:

​$$x^2 + y^2 - 3x - 4y + 1$$​​

​$$1^2 + 2^2 - 3(1) - 4(2) + 1 = 1 + 4 - 3 - 8 + 1 = -5$$​​

Since result < 0, the point lies inside the circle.

Alternate Method:

Formula Used:
The general form of the equation of a circle is:

​$$(x - h)^2 + (y - k)^2 = r^2$$​​

Where (h, k) is the center and r is the radius.

Solution:

Start with the equation of the circle:

​$$x^2 + y^2 - 3x - 4y + 1 = 0$$​​