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Given B is a point of tangency, find the radius of circle C?

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You have given straight line AB tangent to circle at point B on the circle.

It means CB , the radius of the circle is perpendicular to AB.

So We have a right angle triangle ABC ,right angle at B.

By Pythagoras Theorem

AB^2+BC^2=AC^2

But in fig what we have

BC=r (assume)

AC=18+r , r is radius of circle.

and

AB=24

Thus

`(24)^2+r^2=(18+r)^2`

`r^2=(18+r)^2-(24)^2`

`r^2=(18+r-24)(18+r+24)`

`r^2=(r-6)(r+42)`

`r^2=r^2-6r+42r-252`

`0=36r-252`

`36r=252`

`r=252/36`

`r=7 ` unit

Thus radius of the circle will be 7 unit.

First, let r be the radius of the circle. (Refer to the figure below.)

Also, notice that the line segments AB and CB are at right angle. So, we may apply the formula for solving the lengths of the right triangle which is the Pythagorean formula.

`a^2+b^2=c^2`

Base on the figure, the legs of our triangle r and 24. And the hypotenuse is r + 18.

So plug-in a=r, b=24, and c=r+18 to the formula.

`r^2+24^2=(r+18)^2`

Then, expand right side.

`r^2+576=r^2+36+324`

And, then simplify the equation by combining like terms.

So, subtract both sides by r^2.

`r^2-r^2+576=r^2-r^2+36r+324`

`576=36r+324`

Also, subtract both sides by 324.

`576-324=36r+324-324`

`252=36r`

Then, isolate r. To do so, divide both sides by 36.

`252/36=(36r)/36`

`7=r`

**Hence, the radius of the circle is 7.**

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