find the angle between the pair of tangent drawn from (1,3);to the circle x +y-2x+4y-11=0
Answers
Answer:
We have, S=x
2
+y
2
−2x+4y−11=0
And the given point is (1,3)
So S
1
=(1)
2
+(3)
2
−2(1)+4(3)−11=9
and T=x(1)+y(3)−(x+1)+2(y+3)−11=5y−6
So equation of pair of tangent is given by, SS
1
=T
2
⇒9(x
2
+y
2
−2x+4y−11)=(5y−6)
2
⇒9x
2
+9y
2
−18x+36y−99=25y
2
−60y+36
⇒9x
2
−16y
2
−18x+96y−135=0
Comparing with general second degree equation,
a=9,b=−16,h=0
Thus angle between the tangents =tan
−1
Answer:
Step-by-step explanation:
To find the angle between the pair of tangents drawn from (1, 3) to the circle x + y - 2x + 4y - 11 = 0, we can first rewrite the equation of the circle in standard form:x + y - 2x + 4y - 11 = 0
-x + 4y - 11 = 0
y = (11 + x) / 4The center of the circle is at the point (-11/4, 0), and the radius is sqrt((11/4)^2 + 0^2) = sqrt(121) / 4.The equation of the tangent line at the point (1, 3) is of the form y = m(x - 1) + 3, where m is the slope of the tangent line. The slope of the tangent line is equal to the slope of the radius drawn to (1, 3). The slope of the radius drawn to (1, 3) is equal to the difference in y-coordinates between (1, 3) and (-11/4, 0) divided by the difference in x-coordinates between (1, 3) and (-11/4, 0):m = (3 - 0) / (1 - (-11/4)) = 3 / (15/4) = 4/5The equation of the tangent line is therefore given by y = 4/5(x - 1) + 3.The angle between the pair of tangents is equal to the angle between the lines y = 4/5(x - 1) + 3 and y = 0. The angle between two lines is acute if the slope of one line is greater than the slope of the other line. In this case, the slope of the line y = 4/5(x - 1) + 3 is 4/5, and the slope of the line y = 0 is 0. Therefore, the angle between the pair of tangents is acute if 4/5 > 0.Thus, the angle between the pair of tangents drawn from (1, 3) to the circle x + y - 2x + 4y - 11 = 0 is acute.