Circles Test - 21

A tangent to a circle is perpendicular to the radius at the point of tangency.
Using this property and Pythagoras Theorem,
$$AO^2 = AP^2 + OP^2$$
$$5^2 = AP^2 + 4^2$$
Thus, $$AP = 4$$ cm

We use the fact that tangent is perpendicular to radius at the point of contact, so, the distance of the point $$A$$ from the center of the circle is $$\sqrt{3^2+4^2} = 5 cm$$.

We know that, a tangent to a circle is perpendicular to the radius at the point of contact. 

$$ Given-\\ Two\quad concentric\quad circles\quad have\quad the\quad centre\quad as\quad O.\\ AB\quad is\quad a\quad chord\quad of\quad the\quad outer\quad circle\quad and\quad AB\quad touches\quad the\quad inner\\ circle\quad at\quad T.\\ OA=OB=5cm\quad and\quad AB=8cm.\\ To\quad find\quad out-\quad \\ The\quad radius\quad of\quad the\quad inner\quad circle\quad i.e\quad OT=?\\ Solution-\\ AB\quad touches\quad the\quad inner\quad circle\quad at\quad T.\quad \\ So\quad the\quad line\quad OT\quad from\quad the\quad centre\quad O\quad to\quad T\quad is\quad \\ perpendicular\quad to\quad AB\quad at\quad T.\\ i.e\quad \angle OTA={ 90 }^{ o }=\angle OTB.....(i)\\ \therefore \quad Between\quad \Delta OTA\quad \& \quad \Delta OTB\quad we\quad have\quad \\ OA=OB=5cm\quad (given)\\ \angle OTA=\angle OTB={ 90 }^{ o }\quad (from\quad i)\\ \therefore \quad \Delta OTA\quad \& \quad \Delta OTB\quad are\quad congruent\quad by\quad SAS\quad test.\\ \Longrightarrow AT=BT\quad or\quad AT=\frac { 1 }{ 2 } \times AB\\ \Longrightarrow AT=\frac { 1 }{ 2 } \times 8cm=4cm.\\ Now\quad in\quad \Delta OTA\quad we\quad have\quad \angle OTA={ 90 }^{ o }\\ i.e\quad \Delta OTA\quad is\quad right\quad angled\quad at\quad \quad T\quad with\quad OA\quad as\quad hypotenuse.\\ \therefore \quad Applying\quad Pythagoras\quad theorem,\quad we\quad have,\\ { OT }^{ 2 }={ OA }^{ 2 }-{ AT }^{ 2 }\\ \Longrightarrow OT=\sqrt { { OA }^{ 2 }-{ AT }^{ 2 } } =\sqrt { { 5 }^{ 2 }-{ 4 }^{ 2 } } cm=3cm\\ So\quad the\quad radius\quad of\quad the\quad inner\quad circle\quad 3cm.\\ Ans-\quad 3cm.\\  $$

$$ Given-\\ The\quad radii\quad of\quad two\quad concentric\quad cirles\quad with\quad centre\quad O\quad are\quad 5cm\quad \& \quad 4cm.\\ The\quad chord\quad AB\quad of\quad outer\quad circle\quad touches\quad the\quad inner\quad circle\quad at\quad P.\\ To\quad find\quad out-\\ The\quad length\quad of\quad AB=?.\\ $$
$$Solution-\\ AB\quad touches\quad the\quad inner\quad circle\quad at\quad P.\\ \therefore \quad AB=2AP,\quad OP\bot AB\quad \quad \quad \\ \Longrightarrow \Delta AOP\quad is\quad a\quad right\quad angled\quad one\quad with\quad AO\quad as\quad the\quad hypotenuse.\\ Now\quad in\quad \Delta AOP\quad AO=5CM,\quad OP=4cm\\ \therefore \quad AP=\sqrt { { AO }^{ 2 }-{ AP }^{ 2 } } =\sqrt { { 5 }^{ 2 }-{ 4 }^{ 2 } } cm=3cm.\\ \therefore \quad AB=2AP=2\times 3cm=6cm\\ Ans-\quad Option\quad B\\ \\ \\ \\  $$ 

Since tangent and radius are perpendicular at the point of contact, we have: 

Given-

Given $$ PT= 30 \hspace {1 mm} cm$$

We know that, tangent to a circle is perpendicular to radius at the point of contact.