find the equation of the circle which touches both the coordinate axes
and passes through the point (1, 2).
Answers
Can any body answer my question please!!!!
Answer:
solved
Step-by-step explanation:
since the circle touches both the coordinate axes it's centre will be ( ±R ,±R)
the equation will be
(x-R)² + (y-R)² = R²
x² + y² - 2xR - 2yR + R² = 0 , the circle passes through (1,2)
1² + 2² - 6R +R² = 0 , R²-6R+5 = 0 , R = 2 , 3
FOR R = 2
we have 4 cases ; according to the location of the centre
case 1 , centre on 1st quadrant (R,R)
x² + y² - 4x - 4y + 4 = 0
case 2 , centre on 2nd quadrant (-R,R)
x² + y² + 4x - 4y + 4 = 0
case 3 , centre on 3rd quadrant (-R,-R)
x² + y² +4x + 4y + 4 = 0
case 1 , centre on 4th quadrant (R,-R)
x² + y² - 4x + 4y + 4 = 0
FOR R =3
we have 4 cases ; according to the location of the centre
case 1 , centre on 1st quadrant (R,R)
x² + y² - 6x - 6y + 9 = 0
case 2 , centre on 2nd quadrant (-R,R)
x² + y² + 6x - 6y + 9 = 0
case 3 , centre on 3rd quadrant (-R,-R)
x² + y² +6x + 6y + 9 = 0
case 1 , centre on 4th quadrant (R,-R)
x² + y² - 6x + 6y + 9 = 0
conclusion : 8 circles