Tangent and Normal Question5

In this page tangent and normal question5 we are going to see solution of some practice questions from the worksheet.

(5) Find the equations of those tangents to the circle x² + y² = 52, which are parallel to the straight line 2 x + 3 y = 6.

Solution:

First let us draw the rough diagram for the above details

Here the given line is parallel to the tangent line drawn to the circle. So the slopes two lines will be equal.

Slope of the given line 2 x + 3 y = 6

m = - coefficient of x/coefficient of y

= -2/3

Slope of the tangent line drawn to the circle is also m = -2/3  ---- (1)

x² + y² = 52

Slope of the tangent line drawn to the circle

differentiate with respect to x

2 x + 2 y (dy/dx) = 0

2 y (dy/dx) = - 2 x

dy/dx = - 2 x/2 y

dy/dx = - x/y  ---- (2)

(1) = (2)

- 2/3 = - x/y

2 y = 3 x

y = (3 x)/2

now we are going to apply y = 3x/2 in the equation of circle to get the point of contact of the tangent

x² + y² = 52

x² + (3x/2)² = 52

x² + (9 x²/4) = 52

(4 x² + 9 x²)/4 = 52

13 x²/4 = 52

x² = (52 x 4)/13

x² = (4 x 4)

x =  ± 4

x = 4                          x = -4

y = 3(4)/2                   y = 3(-4)/2

y = 6                          y = -6

Therefore the point of contact are (4,6) (-4,-6)

Equation of the tangent passing through the point (4,6) and slope is -2/3

(y - y₁) = m (x - x₁)

(y - 6) = (-2/3) (x - 4)

3 (y - 6) = -2 (x - 4)

3 y - 18 = -2 x + 8

2 x + 3 y - 8 - 18 = 0

2 x + 3 y - 26 = 0

Equation of the tangent passing through the point (-4,-6) and slope is -2/3

(y - y₁) = m (x - x₁)

[y - (-6)] = (-2/3) [x - (- 4)]

[y + 6] = (-2/3) [x + 4]

3 (y + 6) = -2 (x + 4)

3 y + 18 = -2 x - 8

2 x + 3 y + 18 + 8 = 0.

2 x + 3 y + 26 = 0

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