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[tex]$L$[/tex] is the circle with equation \( x^2 + y^2 = 9 \). \( P\left(\frac{3}{2}, \frac{3 \sqrt{3}}{4}\right) \) is a point on \( L \).

a) The gradient of \( OP \) can be written in the form \( \frac{\sqrt{3}}{a} \) where \( a \) is an integer. Work out the value of \( a \).

b) The gradient of the tangent at \( P \) can be written in the form \( \frac{b}{\sqrt{3}} \) where \( b \) is an integer. Work out the value of \( b \).

c) The equation of the tangent to [tex]\( L \)[/tex] at the point [tex]\( P \)[/tex] intersects the [tex]\( y \)[/tex]-axis at [tex]\( \left(0, \frac{7 \sqrt{3}}{c}\right) \)[/tex] where [tex]\( c \)[/tex] is an integer. Work out the value of [tex]\( c \)[/tex].

Sagot :

GinnyAnswer
Let's go through the solution step-by-step.

### Given:
The circle \( L \) with equation \( x^2 + y^2 = 9 \).
Point \( P \left( \frac{3}{2}, \frac{3\sqrt{3}}{4} \right) \) is on the circle.

### Part (a) - The gradient of \( OP \):

The gradient of a line passing through points \( O(0, 0) \) and \( P \left( \frac{3}{2}, \frac{3\sqrt{3}}{4} \right) \) is calculated as follows:
[tex]\[ \text{Gradient of } OP = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{3\sqrt{3}}{4} - 0}{\frac{3}{2} - 0} = \frac{\frac{3\sqrt{3}}{4}}{\frac{3}{2}} \][/tex]

Simplifying the fraction:
[tex]\[ = \frac{3\sqrt{3}}{4} \div \frac{3}{2} = \frac{3\sqrt{3}}{4} \times \frac{2}{3} = \frac{3\sqrt{3} \cdot 2}{3 \cdot 4} = \frac{6\sqrt{3}}{12} = \frac{\sqrt{3}}{2} \][/tex]

So, \( \frac{\sqrt{3}}{a} = \frac{\sqrt{3}}{2} \). Therefore,

[tex]\[ a = 2 \][/tex]

### Part (b) - The gradient of the tangent at \( P \):

The gradient of the tangent to a circle at any point is the negative reciprocal of the gradient of the radius to that point.

The gradient of the tangent at \( P \):
[tex]\[ \text{Gradient of the tangent} = -\frac{1}{\text{Gradient of } OP} = -\frac{1}{\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}} \][/tex]

So, \( \frac{b}{\sqrt{3}} = -\frac{2}{\sqrt{3}} \). Therefore,

[tex]\[ b = -2 \][/tex]

### Part (c) - The equation of the tangent and where it intersects the \( y \)-axis:

The equation of the tangent to the circle at point \( P \left( \frac{3}{2}, \frac{3\sqrt{3}}{4} \right) \) with a gradient of \( -\frac{2}{\sqrt{3}} \) is:

Using the point-slope form \( y - y_1 = m (x - x_1) \):
[tex]\[ y - \frac{3\sqrt{3}}{4} = -\frac{2}{\sqrt{3}} \left( x - \frac{3}{2} \right) \][/tex]

To find the \( y \)-intercept where \( x = 0 \):
[tex]\[ y - \frac{3\sqrt{3}}{4} = -\frac{2}{\sqrt{3}} \left( 0 - \frac{3}{2} \right) \][/tex]
[tex]\[ y - \frac{3\sqrt{3}}{4} = -\frac{2}{\sqrt{3}} \times -\frac{3}{2} \][/tex]
[tex]\[ y - \frac{3\sqrt{3}}{4} = \frac{6}{2\sqrt{3}} \][/tex]
[tex]\[ y - \frac{3\sqrt{3}}{4} = \frac{3\sqrt{3}}{3} \][/tex]
[tex]\[ y - \frac{3\sqrt{3}}{4} = \sqrt{3} \][/tex]
[tex]\[ y = \sqrt{3} + \frac{3\sqrt{3}}{4} \][/tex]
[tex]\[ y = \frac{4\sqrt{3}}{4} + \frac{3\sqrt{3}}{4} \][/tex]
[tex]\[ y = \frac{7\sqrt{3}}{4} \][/tex]

So, the \( y \)-intercept is \( \left( 0, \frac{7\sqrt{3}}{c} \right) \).

Therefore,
[tex]\[ \frac{7\sqrt{3}}{c} = \frac{7\sqrt{3}}{4} \][/tex]

[tex]\[ c = 4 \][/tex]

### Conclusion:

- (a) \( a = 2 \)
- (b) \( b = -2 \)
- (c) [tex]\( c = 4 \)[/tex]
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