Question

find the angle between tangents drawn from (3,2) to circle x2+y2-6x+4y-2=0 explain with the process​

Answer 1

Step-by-step explanation:

Given:

${x}^{2} + {y}^{2} - 6x + 4y - 2 = 0$

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

$\boxed{{x}^{2} + {y}^{2} + 2gx + 2fy + c = 0 }$

here, centre C(-g,-f) and radius r=√(g²+f²-c)

Compare given equation with standard equation

$2gx = - 6x \\ \\ g = - 3 \\ \\ 2fy = 4y \\ \\ f = 2 \\ \\ centre \:\bold{ C= (3 - 2)}$

Radius

$r = \sqrt{(9 + 4 - 2)} \\ \\ r = \sqrt{15}$

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

$\boxed{Distance \: formula = \sqrt{( {x_2 - x_1)}^{2} + ( {y_2 -y_1)}^{2} } }$

P(3,2) and C(3,-2)

$PC = \sqrt{( {3 - 3)}^{2} + ( {2 + 2)}^{2} } \\ \\ \bold{PC = 4 \: units}$

Find Perpendicular by applying Pythagoras theorem,

$PC {}^{2} = {AC}^{2} + {AP}^{2} \\ \\ {AP}^{2} = 16 - 15 \\ \\ \bold{AP = 1 \: units}$

Now,find the angle CPA,by Applying trigonometric ratios on right triangle.

$tan \frac{ \theta}{2} = \frac{AC}{AP} \\ \\ tan \frac{ \theta}{2} = \frac{ \sqrt{15} }{1} \\ \\ \frac{ \theta}{2} = {tan}^{ - 1} ( \sqrt{15} ) \\ \\ \frac{ \theta}{2} = {tan}^{ - 1}(3.87) \\ \\ \frac{ \theta}{2} =75.51° \\ \\ \theta =2\times75.51° \\ \\ \bold{\theta=151.02°}\\\\ \theta\:\approx\:151°$

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

the angle between them

https://brainly.in/question/15732369

Answer 2

Answer:

Angle between pair of tangents is 151°