If tangent drawn from `(2,0)` to the parabola `x=-2y^2` are also tangents to the circle `(x-5)^2+y^2=r^2` then the value of `17r^2` is

Answer

To solve the problem, we need to find the value of \( 17r^2 \) given that the tangents drawn from the point \( (2, 0) \) to the parabola \( x = -2y^2 \) are also tangents to the circle \( (x - 5)^2 + y^2 = r^2 \).

### Step 1: Find the equation of the tangent to the parabola
The equation of the parabola is given by \( x = -2y^2 \). The slope of the tangent line at any point on the parabola can be expressed as \( m \). The equation of the tangent line at a point \( (x_0, y_0) \) on the parabola can be written as:
\[ y - y_0 = m(x - x_0) \]
Substituting \( x_0 = -2y_0^2 \) into the equation, we get:
\[ y - y_0 = m(x + 2y_0^2) \]

### Step 2: Substitute the point from which the tangent is drawn
The tangent is drawn from the point \( (2, 0) \). Therefore, substituting \( (2, 0) \) into the tangent equation gives:
\[ 0 - y_0 = m(2 + 2y_0^2) \]
This simplifies to:
\[ -y_0 = m(2 + 2y_0^2) \]

### Step 3: Express \( y_0 \) in terms of \( m \)
Rearranging the equation gives:
\[ y_0 = -m(2 + 2y_0^2) \]
This can be rewritten as:
\[ y_0 + 2m y_0^2 = -2m \]
This is a quadratic equation in \( y_0 \):
\[ 2m y_0^2 + y_0 + 2m = 0 \]

### Step 4: Find the discriminant for tangency condition
For the tangents to exist, the discriminant of this quadratic equation must be zero:
\[ D = b^2 - 4ac = 1^2 - 4(2m)(2m) = 1 - 16m^2 = 0 \]
Setting the discriminant to zero gives:
\[ 1 - 16m^2 = 0 \implies 16m^2 = 1 \implies m^2 = \frac{1}{16} \implies m = \pm \frac{1}{4} \]

### Step 5: Find the points of tangency
Using \( m = \frac{1}{4} \):
\[ y_0 = -\frac{1}{4}(2 + 2y_0^2) \]
Substituting \( m = \frac{1}{4} \) into the quadratic gives:
\[ 2 \cdot \frac{1}{4} y_0^2 + y_0 + 2 \cdot \frac{1}{4} = 0 \implies \frac{1}{2} y_0^2 + y_0 + \frac{1}{2} = 0 \]
Multiplying through by 2:
\[ y_0^2 + 2y_0 + 1 = 0 \implies (y_0 + 1)^2 = 0 \implies y_0 = -1 \]
Thus, the point of tangency is \( (2, -1) \).

Using \( m = -\frac{1}{4} \):
\[ y_0 = -(-\frac{1}{4})(2 + 2y_0^2) \implies y_0 = \frac{1}{4}(2 + 2y_0^2) \]
Following similar steps leads to the same point of tangency.

### Step 6: Find the distance from the center of the circle to the tangent line
The center of the circle is \( (5, 0) \) and the radius is \( r \). The distance \( d \) from the center of the circle to the tangent line can be calculated using the formula for the distance from a point to a line:
\[ d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For the tangent line \( y = \frac{1}{4}x - \frac{1}{2} \):
\[ A = -\frac{1}{4}, B = 1, C = \frac{1}{2} \]
Substituting \( (x_1, y_1) = (5, 0) \):
\[ d = \frac{|-5/4 + 0 + 1/2|}{\sqrt{(-\frac{1}{4})^2 + 1^2}} = \frac{|-\frac{5}{4} + \frac{2}{4}|}{\sqrt{\frac{1}{16} + 1}} = \frac{|-3/4|}{\sqrt{\frac{17}{16}}} = \frac{3/4}{\frac{\sqrt{17}}{4}} = \frac{3}{\sqrt{17}} \]
Setting this equal to \( r \):
\[ r = \frac{3}{\sqrt{17}} \]

### Step 7: Calculate \( 17r^2 \)
\[ r^2 = \left(\frac{3}{\sqrt{17}}\right)^2 = \frac{9}{17} \]
Thus,
\[ 17r^2 = 17 \times \frac{9}{17} = 9 \]

### Final Answer
The value of \( 17r^2 \) is \( \boxed{9} \).