Equation of a circle passing through the point (1, 2) and (3, 4) and touching the line 3x + y - 3 = 0 is
(1) x2 + y2 - 8x - 2y + 7 = 0
(2) x2 + y2 - 8x – 2y – 7 = 0
(3) x2 + y2 - 4x - 2y + 3 = 0
(4) x2 + y2 - 6x - 4y + 3 = 0
Answers
Given: line 3x + y - 3 = 0, point (1, 2) and (3, 4)
To find: Equation of a circle passing through the point and the line?
Solution:
- Lets consider the centre of the circle be (x,y), so the radius is equal from both the points so:
√(x+1)² + (y-2)² = √(x-3)² + (y-4)² ............(i)
- squaring both sides and expanding we get:
(x-1)² + (y-2)² = (x-3)² + (y-4)²
x² +1 −2x + y²+ 4− 4y = x² + 9 − 6x + y² + 16 − 8y
- cancelling x² and y² from both sides:
−20 + 4x + 4y = 0
x + y = 5
y = 5 − x .................(ii)
- Now, we know that distance from centre is same as that of the radius so:
mod { (3x+y−3) / (3)²+(1)² }
{ 3x = 5 - x - 3 / √10}
mod (2x + 2 )/√10 ................(iii)
- Now put (ii) in (i), and then equate it with the third equation, we get:
√(x+1)² + (3-x)² = mod (2x + 2 )/√10
- squaring both sides
(x+1)² + (3-x)² = (2x + 2 )²/10
x²+1−2x+9+x²−6x=(4x²+4 +8x)/10
10 x (2x²-8x+10) = 4x²+4 +8x
20x²-80x+100 = 4x²+4 +8x
16x² - 88x + 96 = 0
- It can be written in simplest form as:
2x² - 11x + 12 = 0
- for the roots -b±√D / 2a
11 ± √(-11)² - 4(2)(12)/2(2)
11 ± √25/4
11 ± 5 / 4
16/4 , 6/4
4, 3/2.
- Now after substituting x in equation ii, we get y = 1 and 7/2.
- So, two centres are (4,1) and (3/2,7/2).
- So equation of circle is:
(x-4)² + (y-1)² = 10
- After calculation:
x2 + y2 - 8x - 2y + 7 = 0
Answer:
Equation of a circle passing through the point (1, 2) and (3, 4) and touching the line 3x + y - 3 = 0 is x2 + y2 - 8x - 2y + 7 = 0
Answer:
hi
Step-by-step explanation:
answer option 3 is real answer it's image answer key given below