equation of circle touching x axis at origin and the line 4x -3y +24=0 are
Answers
Answer:
x² + y² - 6y = 0
x² + y² + 24y = 0
Step-by-step explanation:
Equation of circle
(x -a)² + (y-b)² = r²
as circle touches x axis at origin
so a = 0 and radius = b
x² + (y - b)² = b²
x² + y² - 2by + b² = b²
x² + y² - 2by = 0
As circle touches line 4x - 3y + 24 = 0
Line from center at touching point (Multiplication of slope of perpendicular lines is -1)
so equation of line
y = -3x/4 + c
for x = 0 , y = b
y = -3x/4 + b
4y = -3x + 4b
3x + 4y - 4b = 0
3x + 4y – 4b = 0 Eq1
4x - 3y + 24 = 0 Eq2
Let’s find their intersection point
3Eq1 + 4Eq2
25x = 12b – 96
X = (12b – 96)/25
4Eq1 - 3*Eq2
25 y = 16b + 72
Y = (16b + 72)/25
This intersection point is at b distance (radius) from center (0,b) point
((12b – 96)/25 – 0)² + ((16b + 72)/25 – b)² = b²
(144/625)(b-8)² + (81/625)(8 –b)² = b²
225(b-8)²= 625b²
9(b-8)²= 25b²
9(b² + 64 – 16b) = 25b²
16b^2 + 144b - 576 = 0
b² + 9b – 36 = 0
b² + 12b – 3b – 36 = 0
b(b+12) -3(b+12) = 0
(b-3)(b+12) = 0
b= 3 & b = -12
x² + y² - 2by = 0
x² + y² - 6y = 0
x² + y² + 24y = 0