Q. 54

Question

Use the Greek method described in Problem 53 to find an equation of the tangent line to the circle

x² + y² - 4x + 6y + 4 = 0 at the point $(3, 2 \sqrt{2} - 3)$

Step-by-Step Solution

Verified

Answer

The equation of the tangent line to the circle is $2 \sqrt{2} y + x = 11 - 6 \sqrt{2} .$

1Step 1. Given information

Here equation of a circle $x^{2} + y^{2} - 4 x + 6 y + 4 = 0 a t t h e p o i n t (3, 2 \sqrt{2} - 3) i s g i v e n .$

We have to find out the equation of the tangent line to the circle .

2Step 2 . Description of finding the center of the circle.

To find center , first convert equation in a standard form , $x^{2} + y^{2} - 4 x + 6 y + 4 = 0 g r o u p t h e x c o n t a i n i n g t e r m a n d y c o n t a i n i n g t e r m (x^{2} - 4 x) + (y^{2} + 6 y) = - 4, n o w c o m p l e t e t h e s q u a r e i n s i d e t h e p a r e n t h e s i s (x^{2} - 4 x + 4) + (y^{2} + 6 y + 9) = - 4 + 4 + 9 {(x - 2)}^{2} + {(y + 3)}^{2} = 9, t h i s i s i n t h e s \tan d a r d f o r m o f e q u a t i o n o f a c i r c l e h e n c e c e n t e r i s (2, - 3)$

3Step 3. Finding the slope

The slope of the line joining center (2,-3) to a given point $(3, 2 \sqrt{2} - 3)$ , then $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} = \frac{2 \sqrt{2} - 3 + 3}{3 - 2} = 2 \sqrt{2} m_{2} = \frac{- 1}{m} \sin c e i t i s p e r p e n d i c u l a r t o l i n e j o i n i n g c e n t e r = \frac{- 1}{2 \sqrt{2}} t h i s s l o p e i s l i n e p a s \sin g t h r o u g h t h e p o i n t (3, 2 \sqrt{2} - 3) .$

4Step 4. Description of finding equation of the tangent line to the circle

The equation of a line passing through (x₁,y₁ ) and slop m is $(y - y_{1}) = m (x - x_{1}) (y - (2 \sqrt{2} - 3)) = \frac{- 1}{2 \sqrt{2}} (x - 3), h e r e (x_{1,} y_{1}) i s (3, 2 \sqrt{2} - 3) a m d m = \frac{- 1}{2 \sqrt{2}} y - 2 \sqrt{2} + 3 = \frac{- 1}{2 \sqrt{2}} (x - 3) (y - 2 \sqrt{2} + 3) \times 2 \sqrt{2} = - x + 3 2 \sqrt{2} y - 8 + 6 \sqrt{2} = - x + 3 2 \sqrt{2} y + x = 11 - 6 \sqrt{2}, t h i s i s t h e e q u a t i o n o f t h e \tan g e n t l i n e t o t h e c i r c l e .$

Q.53

Q.55

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Q.56

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