Find the pair of tangents drawn from P(3,2) to the circle x² + y² - 6x + 4y - 2 = 0
Answers
Step-by-step explanation:
Given circle is S = x² + y² - 6xy + 4y - 2 = 0 and point P(3,2)
Pair of tangents to the circle S = 0 drawn from P(3,2) is SS₁₁ = S₁²
S₁ = x(3) + y(2) - 3(x + 3) + 2(y + 2) - 2
∵ xy₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
⇒ 3x + 2y - 3x - 9 + 2y + 4 - 2
⇒ S₁ = 4y - 7
and S₁₁ = (3)² + (2)² - 6(3) + 4(2) - 2 = 1
Equation of pair of tangents is SS₁₁ = S₁²
⇒ (x² + y² - 6x + 4y - 2)(1) = (4y - 7)²
∵ x₁² + y₁² + 2gx₁ + 2fy₁ + c = 0
⇒ 16y² + 49 - 56y = x² + y² - 6x + 4y - 2
⇒ x² - 15y² - 6x + 60y - 51 = 0
Answer:
x
2
+y
2
−6x+4y−2=0
Point (3,2)
To find:find the angle between tangents drawn from (3,2) to circle x²+y²-6x+4y-2=0
Solution:
Find the centre and radius of circle.
we know that standard equation of circle is
\begin{gathered} \boxed{{x}^{2} + {y}^{2} + 2gx + 2fy + c = 0 }\\ \\ \end{gathered}
x
2
+y
2
+2gx+2fy+c=0
here, centre C(-g,-f) and radius r=√(g²+f²-c)
Compare given equation with standard equation
\begin{gathered}2gx = - 6x \\ \\ g = - 3 \\ \\ 2fy = 4y \\ \\ f = 2 \\ \\ centre \:\bold{ C= (3 - 2)} \\ \\ \end{gathered}
2gx=−6x
g=−3
2fy=4y
f=2
centreC=(3−2)
Radius
\begin{gathered}r = \sqrt{(9 + 4 - 2)} \\ \\ r = \sqrt{15} \\ \\ \end{gathered}
r=
(9+4−2)
r=
15
See the attached figure, since pair of tangents can be drawn from a single point.
Tangent; at the point of contact , makes 90° angle with radius of circle.
∆CPA is right angle at A.
Since CA is radius= √15 units
PC: Find distance from distance formula
\begin{gathered}\boxed{Distance \: formula = \sqrt{( {x_2 - x_1)}^{2} + ( {y_2 -y_1)}^{2} } } \\ \\ \end{gathered}
Distanceformula=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
P(3,2) and C(3,-2)
\begin{gathered}PC = \sqrt{( {3 - 3)}^{2} + ( {2 + 2)}^{2} } \\ \\ \bold{PC = 4 \: units} \\ \end{gathered}
PC=
(3−3)
2
+(2+2)
2
PC=4units
Find Perpendicular by applying Pythagoras theorem,
\begin{gathered}PC {}^{2} = {AC}^{2} + {AP}^{2} \\ \\ {AP}^{2} = 16 - 15 \\ \\ \bold{AP = 1 \: units} \\ \\ \end{gathered}
PC
2
=AC
2
+AP
2
AP
2
=16−15
AP=1units
Now,find the angle CPA,by Applying trigonometric ratios on right triangle.
\begin{gathered}tan \frac{ \theta}{2} = \frac{AC}{AP} \\ \\ tan \frac{ \theta}{2} = \frac{ \sqrt{15} }{1} \\ \\ \frac{ \theta}{2} = {tan}^{ - 1} ( \sqrt{15} ) \\ \\ \frac{ \theta}{2} = {tan}^{ - 1}(3.87) \\ \\ \frac{ \theta}{2} =75.51° \\ \\ \theta =2\times75.51° \\ \\ \bold{\theta=151.02°}\\\\ \theta\:\approx\:151° \end{gathered}
tan
2
θ
=
AP
AC
tan
2
θ
=
1
15
2
θ
=tan
−1
(
15
)
2
θ
=tan
−1
(3.87)
2
θ
=75.51°
θ=2×75.51°
θ=151.02°
θ≈151°
Thus,
Angle between pair of tangents is 151°.
Hope it helps you.
To learn more on brainly:
Find the pair of tangents drawn from (1,3) to the circle x² + y2 - 2x + 4y -11=0 and also find