#### To determine

**To find:**

The smallest value of r such that any line with slope 25 intersects some of these circles

**Solution:**The smallest value of r such that any line with slope 25 intersects some of these circles is, r≈0.093

#### Explanation

**1) Formula:**

i. Tangent normal formula: If m1 is the slope of tangent and m2 is the slope of normal then,

m1*m2=-1

ii. Equation of line passing through the origin is,

y=mx

iii. Point-slope formula:

m=y-y1x-x1

**2) Given:**

Slope of Tangent line is 25

3) **Calculation:**

By using tangent normal formula, the slope of the normal line is,

m1*m2=-1

⇒25m2=-1

⇒ m2=-52

Equation of line passing through the origin is, y=mx

Therefore,

y=-52x

Now, equation of circle centered at origin is, x2+y2=r2

Solving these equations,

x2+-52x2=r2

⇒x2+254x2=r2

⇒294x2=r2

⇒x2=429r2

⇒x=229r

Substitute in y=-52x

Therefore,

y=-52229r=-529r

Because of periodic nature of lattice points, it suffices to consider points in 5/2 grid shown. We can see that the minimum value of r occurs when there is a line with slope 2 which touches the circle which centered at (3, 1) and circle centered at (0,0) and (5, 2)

To find point Q, we solve x-32+y-12=r2 and y-1= -52(x-3)

As above we get,

x=3-229r and y=1+529r

Now slope of the line PQ is 25

Therefore, by using point-slope formula

m=y-y1x-x1

⇒25=1+529r+529r3-229r-229r

⇒25=1+1029r3-429r

⇒2(3-429r)=5(1+1029r)

⇒6-829r=5+5029r

⇒5029r+829r=6-5

⇒5829r=1

⇒r=2958≈0.093

**Conclusion:**

The smallest value of r such that any line with slope 25 intersects some of these circles is,

r≈0.093