Find the equation of the circle which touches the y-axis and whose centre is (2, -3). Find the equation of the circle x² + y² - 8x + 14y + 1 = 0 and has half of its area.

Steps to Solve

1. Identify the radius of the first circle

The center of the circle is given as $(2, -3)$. Since the circle touches the y-axis, the radius (r) is equal to the x-coordinate of the center. Thus,

$$ r = 2 $$

2. Write the equation of the first circle

Using the center-radius form of a circle's equation, the equation is:

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

Substituting the values (h = 2), (k = -3), and (r = 2):

$$(x - 2)^2 + (y + 3)^2 = 2^2$$

This simplifies to:

$$(x - 2)^2 + (y + 3)^2 = 4$$

3. Convert the given circle's equation to standard form

The given equation is:

$$x^2 + y^2 - 8x + 14y + 1 = 0$$

We rearrange and complete the squares for (x) and (y):

Group the terms:

$$ (x^2 - 8x) + (y^2 + 14y) + 1 = 0 $$

Completing the square:

For (x): $$ x^2 - 8x = (x - 4)^2 - 16 $$

For (y): $$ y^2 + 14y = (y + 7)^2 - 49 $$

Substituting back, we have:

$$ (x - 4)^2 - 16 + (y + 7)^2 - 49 + 1 = 0 $$

Simplifying further leads to:

$$ (x - 4)^2 + (y + 7)^2 = 64 $$

Thus, this circle's center is ( (4, -7) ) and the radius is ( r = 8 ).

4. Calculate the area and find half of it

The area (A) of the original circle is:

$$ A = \pi r^2 = \pi (8)^2 = 64\pi $$

Half of this area is:

$$ \frac{A}{2} = \frac{64\pi}{2} = 32\pi $$

5. Find the radius for the new circle

Let (R) be the radius of the new circle, and set its area equal to (32\pi):

$$ \pi R^2 = 32\pi $$

Dividing both sides by (\pi):

$$ R^2 = 32 $$

Taking the square root:

$$ R = 4\sqrt{2} $$

6. Write the equation of the new concentric circle

Since it is concentric (same center as the previous circle at ( (4, -7) )), the equation for the new circle is:

$$(x - 4)^2 + (y + 7)^2 = (4\sqrt{2})^2$$

This simplifies to:

$$(x - 4)^2 + (y + 7)^2 = 32$$