Question Statement

Two tangents are drawn from a point $P(6,1)$ lying outside the circle $x^2+y^2-8x-2y+14=0$. Find the area of the shaded region.

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Background and Explanation

To solve this problem, we need to understand the properties of circles and tangents. A tangent is perpendicular to the radius at the point of contact, forming a right-angled triangle. We use the general equation of a circle to find its center and radius, then apply trigonometry and area formulas for triangles and sectors to find the specific region requested.

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Solution

Step 1: Find the Center and Radius of the Circle

The given equation of the circle is: $x^2+y^2-8x-2y+14=0$

Comparing this to the general form $x^2+y^2+2gx+2fy+c=0$, we find:

• The center $C$ is $\left(\frac{-8}{-2},\frac{-2}{-2}\right)=C(4,1)$.
• The radius $r$ is calculated as: $r=\sqrt{g^2+f^2-c}=\sqrt{4^2+1^2-14}=\sqrt{16+1-14}=\sqrt{3}$

Step 2: Calculate the Distance from Point P to the Center

The distance between $P(6,1)$ and $C(4,1)$ is: $|PC|=\sqrt{(6-4{)}^2+(1-1{)}^2}=\sqrt{2^2+0}=\sqrt{4}=2$

Step 3: Determine the Angle $\alpha$

In the right-angled triangle $APC$ (where the angle at $A$ is $90^{\circ }$ because the radius is perpendicular to the tangent): $\cos \alpha =\frac{|\overline{AC}|}{|\overline{PC}|}$ $\cos \alpha =\frac{r}{2}=\frac{\sqrt{3}}{2}$ $\alpha ={\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)=30^{\circ }=\frac{\pi }{6}\text{ radians}$

Step 4: Calculate the Area of Sector $CDA$

The area of the sector is given by $\frac{1}{2}{r}^2\alpha$: $\text{Area of sector }CDA=\frac{1}{2}(\sqrt{3}{)}^2\cdot \frac{\pi }{6}=\frac{1}{2}(3)\cdot \frac{\pi }{6}=\frac{\pi }{4}\text{ sq. units}$

Step 5: Calculate the Length of Tangent $AP$ and Area of $△ACP$

Using trigonometry in $△ACP$: $\frac{|\overline{AP}|}{|\overline{AC}|}=\tan \alpha ⟹|AP|=\sqrt{3}\tan 30^{\circ }=\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)=1$

Now, the area of the right-angled triangle $ACP$ is: $\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}|\overline{AC}||\overline{AP}|$ $\text{Area}=\frac{1}{2}(\sqrt{3})(1)=\frac{\sqrt{3}}{2}\text{ sq. units}$

Step 6: Calculate the Total Shaded Area

The shaded region $PCA$ (one half of the total shaded area) is the difference between the triangle and the sector: $\text{Area of region }PCA=\text{Area of }△ACP-\text{Area of sector }CDA=\frac{\sqrt{3}}{2}-\frac{\pi }{4}$

Since the diagram is symmetric, the total shaded area is twice the area of $PCA$: $\text{Total Shaded Area}=2\times \left(\frac{\sqrt{3}}{2}-\frac{\pi }{4}\right)$ $\text{Total Shaded Area}=\sqrt{3}-\frac{\pi }{2}\text{ sq. units}$

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Key Formulas or Methods Used

• General Circle Equation: $x^2+y^2+2gx+2fy+c=0$ with Center $(-g,-f)$ and Radius $\sqrt{g^2+f^2-c}$.
• Distance Formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$.
• Trigonometric Ratios: $\cos \theta =\frac{\text{adj}}{\text{hyp}}$ and $\tan \theta =\frac{\text{opp}}{\text{adj}}$.
• Area of a Sector: $A=\frac{1}{2}{r}^2\theta$ (where $\theta$ is in radians).
• Area of a Triangle: $A=\frac{1}{2}\times \text{base}\times \text{height}$.

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Summary of Steps

1. Identify the center $(4,1)$ and radius $\sqrt{3}$ from the circle's equation.
2. Calculate the distance between the external point $P$ and the center $C$.
3. Use the cosine ratio in the right triangle formed by the tangent and radius to find the internal angle $\alpha =30^{\circ }$.
4. Calculate the area of the circular sector using the radius and angle.
5. Calculate the area of the right triangle $ACP$.
6. Subtract the sector area from the triangle area to find the upper shaded part.
7. Double the result to account for the symmetric lower portion of the shaded region.