9. Write down the centre and radius length of the circle $$ x^{2}+y^{2}-6 x-2 y-15=0 $$. If the line $$ 3 x+4 y+c=0 $$ is a tangent to this circle, find the two values of $$ c $$. 10. For what values of $$ k $$ is the line $$ 2 x-k y-3=0 $$ a tangent to the circle

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**Problem 9**

We are given the circle  
$$
x^{2}+y^{2}-6x-2y-15=0.
$$
  
1. **Finding the centre and radius:**  
   Complete the square in $$ x $$ and $$ y $$.

- For $$ x $$:  
     $$
     x^2 - 6x = \Bigl(x^2 - 6x + 9\Bigr) - 9 = (x-3)^2 - 9.
     $$
     
   - For $$ y $$:  
     $$
     y^2 - 2y = \Bigl(y^2 - 2y + 1\Bigr) - 1 = (y-1)^2 - 1.
     $$
     
   Substitute these into the equation:
   $$
   (x-3)^2 - 9 + (y-1)^2 - 1 - 15 = 0.
   $$
   Combine the constant terms:
   $$
   (x-3)^2 + (y-1)^2 - 25 = 0 \quad \Longrightarrow \quad (x-3)^2 + (y-1)^2 = 25.
   $$
  
   Thus, the centre of the circle is  
   $$
   (3,1)
   $$
   and the radius is  
   $$
   5.
   $$

2. **Finding the values of $$ c $$ for the tangent line $$ 3x+4y+c=0 $$:**  
   A line is tangent to a circle if the distance from the centre to the line equals the radius. For the line  
   $$
   3x+4y+c=0,
   $$
   the distance $$ d $$ from the centre $$ (3,1) $$ is given by  
   $$
   d = \frac{|3(3)+4(1)+c|}{\sqrt{3^2+4^2}} = \frac{|9+4+c|}{5} = \frac{|13+c|}{5}.
   $$
   Setting the distance equal to the radius $$ 5 $$:
   $$
   \frac{|13+c|}{5} = 5.
   $$
   Multiply both sides by 5:
   $$
   |13+c|=25.
   $$
   This absolute value equation leads to two cases:

- Case 1:  
     $$
     13+c=25 \quad \Longrightarrow \quad c=25-13=12.
     $$
     
   - Case 2:  
     $$
     13+c=-25 \quad \Longrightarrow \quad c=-25-13=-38.
     $$
     
   Therefore, the two values of $$ c $$ are  
   $$\displaystyle 12 $$ and $$\displaystyle -38$$.

---

**Problem 10**

We are to find the values of $$ k $$ for which the line  
$$
2x - ky - 3 = 0
$$
is tangent to the same circle  
$$
x^{2}+y^{2}-6x-2y-15=0.
$$
  
1. **Using the distance from the centre to the line:**  
   The centre of the circle is $$ (3,1) $$ and its radius is $$ 5 $$. The distance $$ d $$ from the centre $$ (3,1) $$ to the line $$ 2x-ky-3=0 $$ is:
   $$
   d = \frac{|2(3) - k(1) - 3|}{\sqrt{2^2+k^2}} = \frac{|6 - k - 3|}{\sqrt{4+k^2}} = \frac{|3-k|}{\sqrt{4+k^2}}.
   $$
   Setting the distance equal to $$ 5 $$:
   $$
   \frac{|3-k|}{\sqrt{4+k^2}} = 5.
   $$

2. **Solving the equation:**  
   Multiply both sides by $$\sqrt{4+k^2}$$:
   $$
   |3-k| = 5\sqrt{4+k^2}.
   $$
   Square both sides to remove the absolute value:
   $$
   (3-k)^2 = 25(4+k^2).
   $$
   Expand and simplify:
   $$
   9 - 6k + k^2 = 100 + 25k^2.
   $$
   Bring all terms to one side:
   $$
   9 - 6k + k^2 - 100 - 25k^2 = 0 \quad \Longrightarrow \quad -6k - 24k^2 - 91 = 0.
   $$
   Multiply the entire equation by $$-1$$:
   $$
   24k^2 + 6k + 91 = 0.
   $$
  
3. **Checking for real solutions:**  
   The quadratic equation in $$ k $$ is  
   $$
   24k^2 + 6k + 91 = 0.
   $$
   Compute the discriminant:
   $$
   \Delta = 6^2 - 4\cdot24\cdot91 = 36 - 8736 = -8700.
   $$
   Since $$\Delta < 0$$, there are no real solutions for $$ k $$.

Thus, there is no real value of $$ k $$ for which the line $$ 2x - ky - 3 = 0 $$ is tangent to the circle.
**Problem 9** - **Circle Equation:** $$ x^{2}+y^{2}-6x-2y-15=0 $$ - **Centre:** $$(3, 1)$$ - **Radius:** $$5$$ - **Tangent Line:** $$3x + 4y + c = 0$$ - **Values of $$ c $$:** $$12$$ and $$-38$$ **Problem 10** - **Line Equation:** $$2x - ky - 3 = 0$$ - **Values of $$ k $$:** No real values of $$ k $$ make the line tangent to the circle.

Write down the centre and radius length of the circle .
If the line is a tangent to this circle, find the two values of .

For what values of is the line a tangent to the circle

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Write down the centre and radius length of the circle . If the line is a tangent to this circle, find the two values of . For what values of is the line a tangent to the circle