Question

A circle is tangent to the line given by the equation 3x-y= 15 at the point (9,12). The center of the radius falls on the line given by 2y+x=8. Find the equation of the circle and identify its center and radius

Answer 1

The equation of the circle is x² + y² + 12x - 14y - 165 = 0 , its center is (-6, 7) and its radius is 5√10 units.

Explanation:

a. The center of the circle

Since the equation of the tangent of the circle is 3x - y = 15, we write it in standard form. So y = -3x - 15. Its gradient is -3. The perpendicular line tot he tangent is the radius and it passes through the center of the circle and the point (9, 12). For two perpendicular lines of gradient m₁ and m₂, m₁m₂ = -1. If m₁ = -3, then m₂ the gradient of our radius line is gotten from m₁m₂ = -1

m₂ = -1/m₁ = -1/-3 = 1/3

Since the radius also passes through the point (9, 12), its equation is gotten from

(y - y₁)/(x - x₁) = m₂ where (x₁, y₁) = (9, 12)

Substituting the value of the variables into the equation, we have

So, (y - 12)/(x - 9) = 1/3

cross-multiplying, we have

3(y - 12) = (x - 9)

expanding the brackets, we have

3y - 36 = x - 9

collecting like terms, we have

3y = x - 9 + 36

3y = x + 27

So, the equation of the radius line is

3y - x = 27.

Now, since the center of the circle lies on the line 2y + x = 8, the radius line and this line intersect at the center of the circle. So, we solve both equations simultaneously to find the center of the circle.

So, 3y - x = 27 (1)

2y + x = 8 (2)

adding (1) and (2), we have

5y = 35

y = 35/5

y = 7

Substituting y = 7 into (2), we have

2y + x = 8

2(7) + x = 8

14 + x = 8

x = 8 - 14

x = -6

So, the center of the circle is at (-6, 7)

b. The radius of the circle

The radius of the circle is the length of the line from the tangent point to the center of the circle. So, r = √[(x₂ - x₁)² + (y₂ - y₁)²] where (x₁, y₁) = (-6, 7) and (x₂, y₂) = (9, 12).

Substituting these into r, we have

r = √[(9 - (-6))² + (12 - 7)²]

= √[(9 + 6)² + (12 - 7)²]

= √[15² + 5²]

= √[225 + 25]

= √250

= √25 × √10

= 5√10 units

c. The equation of the circle

Now, the equation of a circle with center (h,k) and radius, r is

(x - h)² + (y - k)² = r² where (h. k) = (-6,7)

Substituting the variables into the equation, we have

(x - (-6))² + (y - 7)² = (√250)²

(x + 6)² + (y - 7)² = 250

expanding the brackets, we have

x² + 12x + 36 + y² - 14y + 49 = 250

collecting like terms, we have

x² + y² + 12x - 14y + 36 + 49 - 250 = 0

x² + y² + 12x - 14y - 165 = 0

So, the equation of the circle is x² + y² + 12x - 14y - 165 = 0 , its center is (-6, 7) and its radius is 5√10 units.