Question

The circle with the centre (4,-1) and touching x axis

Answer 1

Answer:

$x^{2} + y^{2} - 8x + 2y + 16 = 0$

Step-by-step explanation:

The centre of given circle = (4, -1)

The circle is touching the X-axis.

Then the equation of the circle-

$(x-h)^{2}+(y-k)^{2} = r^{2}$

Where (h, k) = (4, -1)

r = radius of circle

Since the circle is touching the X-axis. Then

r = k

r = -1

So, $(x-4)^{2}+(y+1)^{2} = (-1)^{2}$

$(x^{2} - 8x + 16) + (y^{2} + 1 + 2y) = 1$

$x^{2} + y^{2} - 8x + 2y + 17 = 1$

$x^{2} + y^{2} - 8x + 2y + 16 = 0$

This is the required equation of circle.