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Question Number 183962 by Acem last updated on 01/Jan/23

Commented by Acem last updated on 01/Jan/23

$$TBisatangenttothecircle$$

Answered by mr W last updated on 01/Jan/23

$$\left(a\right)$$$$\frac{\sqrt{8^2-{TM}^2}}{8}=\frac{8}{10}$$$$\Rightarrow TM=\frac{24}{5}$$$$ON=\frac{TM}{2}=\frac{12}{5}$$$$\left(b\right)$$$$yes,withOBasdiameterand$$$$centeratmidpointofOB.$$$$radiusr=\frac{OB}{2}=\frac{\sqrt{8^2+3^2}}{2}=\frac{\sqrt{73}}{2}$$

Commented by Acem last updated on 02/Jan/23

$$ThanksSir!$$

Answered by Acem last updated on 02/Jan/23

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