length of tangent drawn from point (5,-2) to point of tangecy on circle x^2+y^2-3x+5y=0?

Question

i need step by step answers please.

Answer 1

The circle is x ²+ y ² - 3 x + 5 y = 0 and the point is (5, - 2).

x ²+ y ² - 3 x + 5 y = 0

x ²- 2 x * 3/2 + (3/2)²- (3/2)² + y ²  + 2 y * 5/2 + (5/2)²- (5/2)² = 0

x ²- 2 x * 3/2 + (3/2)² + y ²  + 2 y * 5/2 + (5/2)² = (3/2)² + (5/2)²

x ²- 2 x * 3/2 + (3/2)² + y ²  + 2 y * 5/2 + (5/2)² = (9/4) + (25/4)

( x - 3 / 2 ) ²+ (y + 5 / 2 ) ² = 36 / 4 

( x - 3 / 2 ) ²+  [ y- (- 5 / 2 ) ] ² = 9

( x - 3 / 2 ) ²+  [ y- (- 5 / 2 ) ] ² = 3 ²

Comparing the circle equation ( x - h ) ²+ ( y - k) ² = r ²

center = (h,k) = (3/2,- 5/2) and  radius = 3.

Let's check the distance between (3/2,- 5/2) and  given point (5, -2) 

Distance

units

3.5355 > 3, so the point (5, - 2) is outside the circle.

Radius of the circle is 3 . This will be one leg of a right triangle.

Distance from the given point to the center of the circle is 3.53 units. This will be the hypotenuse of the right triangle.

 Using the Pythagorean theorem, find the length of the other leg of the triangle.

(hypotenuse) ² = (side) ² + (side) ²  

( 3.53 ) ² = (3) ² + (side) ² 

12.46 - (3) ² = (side) ²

 12.46 - 9 = (side) ² 

3. 46 = (side) ²

side = √ (3.46) = 1.87

side = 1.87 units.

So 1.87units is the length of tangent drawn from point (5, - 2) to point of tangecy on circle.