Show that the line 7x-3y-1=0 touches the circle x²+y²-3x+2y=0
Answers
Answer:
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Step-by-step explanation:
Line 7x - 3y - 1 = 0 touches the circle x² + y² + 5x - 7y + 4 = 0 at the point (1, 2).
Step-by-step explanation:
Equation of the line is given as 7x - 3y - 1 = 0 ---------(1)
and the circle is x² + y² + 5x - 7y + 4 = 0 ------(2)
From equation (1),
y = \frac{1}{3}(7x-1)
3
1
(7x−1)
Now we place the value of y in the equation of circle to get the common point where the line and circle touch each other.
x² + [\frac{1}{3}(7x-1)
3
1
(7x−1) ² + 5x - 7[
x² + \frac{1}{9}(7x-1)^{2}
9
1
(7x−1)
2
+ 5x -\frac{7}{3}(7x-1)
3
7
(7x−1) + 4 = 0
9x² + (49x² + 1 - 14x) + 45x - 21(7x - 1) + 4 = 0
9x² + 49x² + 1 - 14x + 45x - 147x + 21 + 4 = 0
58x² - 116x + 58 = 0
x² - 2x + 1 = 0
(x - 1)² = 0
x = 1
From equation 1,
y = 2
We get only one solution (1, 2) by the solving the equations that means line doesn't intersect the circle, it just touches the circle at (1, 2).
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Answer:
Step-by-step explanation:
Equation of the line is given as 7x - 3y - 1 = 0 ---------(1)
and the circle is x² + y² + 5x - 7y + 4 = 0 ------(2)
From equation (1),
y =
Now we place the value of y in the equation of circle to get the common point where the line and circle touch each other.
x² + []² + 5x - 7[
x² + + 5x - + 4 = 0
9x² + (49x² + 1 - 14x) + 45x - 21(7x - 1) + 4 = 0
9x² + 49x² + 1 - 14x + 45x - 147x + 21 + 4 = 0
58x² - 116x + 58 = 0
x² - 2x + 1 = 0
(x - 1)² = 0
x = 1
From equation 1,
y = 2
We get only one solution (1, 2) by the solving the equations that means line doesn't intersect the circle, it just touches the circle at (1, 2).