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Find the equation of the circle in its general form if its center is the point C(2,-2) and the line 10x - 12y + 18 = 0, is tangent to the circle

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Answer to a math question Find the equation of the circle in its general form if its center is the point C(2,-2) and the line 10x - 12y + 18 = 0, is tangent to the circle

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Birdie
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Solution:

1. Determine the distance from the center of the circle C(2, -2) to the line 10x - 12y + 18 = 0 , which is the radius of the circle:

- The formula for distance from a point (x_1, y_1) to a line ax + by + c = 0 is:

\frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}}

- Substitute a = 10 , b = -12 , c = 18 , and the center (x_1, y_1) = (2, -2) :

D = \frac{|10(2) - 12(-2) + 18|}{\sqrt{10^2 + (-12)^2}} = \frac{|20 + 24 + 18|}{\sqrt{100 + 144}} = \frac{62}{\sqrt{244}}

2. Simplify the denominator:

- \sqrt{244} = \sqrt{4 \times 61} = 2\sqrt{61}

- So, D = \frac{62}{2\sqrt{61}} = \frac{31}{\sqrt{61}}

3. The radius r = \frac{31}{\sqrt{61}} .

4. Write the equation of the circle in the standard form:

- The equation of a circle with center (h, k) and radius r is:

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

- Substitute h = 2 , k = -2 , and r = \frac{31}{\sqrt{61}} :

(x - 2)^2 + (y + 2)^2 = \left(\frac{31}{\sqrt{61}}\right)^2

5. Calculate r^2 :

- \left(\frac{31}{\sqrt{61}}\right)^2 = \frac{31^2}{61} = \frac{961}{61}

6. Expand the circle equation into general form:

- (x - 2)^2 + (y + 2)^2 = \frac{961}{61}

- Expand: x^2 - 4x + 4 + y^2 + 4y + 4 = \frac{961}{61}

7. Multiply everything by 61 to clear the fraction:

- 61x^2 - 244x + 244 + 61y^2 + 244y + 244 = 961

8. Write the general form of the circle equation:

- 61x^2+61y^2-244x+244y-473=0

-

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