Radius OA of circle O has a slope of 2.5. What is the slope of the line tangent to circle O to point A?

2 answers:

5 0

The tangent at A is perpendicular to OA, so has a slope that is the negative reciprocal of that of the radius: -1/2.5 = -2/5.

Mekhanik [1.2K]3 years ago

4 0

Answer: The required slope of the line tangent to the circle at point A is -0.4.

Step-by-step explanation: Given that the radius OA of a circle with center O has a slope of 2.5.

We are to find the slope of the line tangent to the circle at the point A.

We know that

The radius of a circle at a point on the circumference of a circle is perpendicular to the tangent line at the same point on the circumference of the circle.

Also, the product of the slopes of two perpendicular lines is -1.

So, if m is the slope of the tangent line at the point A, then we must have

$m\times 2.5=-1\\\\\Rightarrow m=-\dfrac{1}{2.5}\\\\\\\Rightarrow m=-\dfrac{10}{25}\\\\\Rightarrow m=-\dfrac{2}{5}\\\\\Rightarrow m=-0.4.$

Thus, the required slope of the line tangent to the circle at point A is -0.4.

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√180 can be expressed in the form p√q. where p and q are integers. Find the smallest value of p + q.

frutty [35]

Answer:

The smallest value of p+q is 11

It happens when p = 6 and q = 5.

=======================================================

Explanation:

Let's factor 180 in such a way that exactly one factor is a perfect square.

I'll ignore the trivial factor of 1.

Here are the possible factorizations we could go with:

180 = 4*45

180 = 9*20

180 = 36*5

Those factorizations then lead to the following

$\sqrt{180} = \sqrt{4*45} = \sqrt{4}*\sqrt{45}= 2\sqrt{45}\\\\\sqrt{180} = \sqrt{9*20} = \sqrt{9}*\sqrt{20}= 3\sqrt{20}\\\\\sqrt{180} = \sqrt{36*5} = \sqrt{36}*\sqrt{5}= 6\sqrt{5}\\\\$