Circle
Category : 10th Class
Circle
Circle
(i) There is no tangent passing through a point lying inside the circle. |
(ii) There is one and only one tangent passing through a point lying on the circle. |
(iii) There are exactly two tangents through a point lying outside a circle. |
(a) 7 cm (b) 3 cm
(c) 5 cm (d) 2 cm
(e) None of these
Ans. (b)
Explanation: We know that the lengths of tangents to a circle from an external point are equal.
Let \[AD\text{ }=\text{ }AF\text{ }=\text{ }x\text{ }cm,\text{ }BD\text{ }=\text{ }BE\text{ }=\text{ }y\text{ }cm\text{ }and\text{ }CE\text{ }=\text{ }CF\text{ }=\text{ }z\text{ }cm\]
Now, \[AB\text{ }=\text{ }12\,cm\] \[\Rightarrow \] \[AD\text{ }+\text{ }BD\text{ }=\text{ }12\text{ }cm\] \[\Rightarrow \] \[x\text{ }+\text{ }y\text{ }=\text{ }12\text{ }cm\] ..... (i)
\[BC\text{ }=\text{ }8\text{ }cm\] \[\Rightarrow \] \[BE\text{ }+\text{ }CE\text{ }=\text{ }8\text{ }cm\] \[\Rightarrow \] \[y\text{ }+\text{ }z\text{ }=\text{ }8\text{ }cm\] .... (ii)
and \[AC\text{ }=\text{ }10\text{ }cm\]\[\Rightarrow \] \[AF\text{ }+\text{ }CF\text{ }=\text{ }10\text{ }cm\] \[\Rightarrow \] \[x\text{ }+\text{ }z\text{ }=\text{ }10\text{ }cm\] .... (iii)
Adding (i), (ii) and (iii) we get:
\[2\left( x\text{ }+\text{ }y\text{ }+\text{ }z \right)=30\] \[\Rightarrow \] \[x\text{ }+\text{ }y\text{ }+\text{ }z\text{ }=\text{ }15\] ...(iv)
From (i) and (iv) we get: \[z\text{ }=\text{ }3\text{ }cm\].
2. In the following figure, AB and CD are two common tangents to the two touching circles. If DC = 4 cm then find the length of AB.
(a) 2 cm (b) 8 cm
(c) 10 cm (d) 12 cm
(e) None of these
Ans. (b)
Explanation: We know that the lengths of two tangents drawn to a circle from a point outside the circle are equal.
So, \[DA\text{ }=\text{ }DC\text{ }and\text{ }DB\text{ }=\text{ }DC\]
\[\therefore \] \[AB\text{ }=\text{ }DA\text{ }+\text{ }DB\text{ }=\text{ }2DC\text{ }=\text{ }8\text{ }cm\]
(a) \[AB\text{ }+\text{ }CD\text{ }=\text{ }BC\text{ }+\text{ }AD\]
(b) \[AC\text{ }+\text{ }AD\text{ }=\text{ }BD\text{ }+\text{ }CD\]
(c) \[AB\text{ }+\text{ }CD\text{ }=\text{ }AC\text{ }+\text{ }BC\]
(d) \[AC\text{ }+\text{ }AD\text{ }=\text{ }BC\text{ }+\text{ }DB\]
(e) None of these
Ans. (a)
Explanation: We know that the tangents to a circle from a point outside it are equal.
So, \[AP\text{ }=\text{ }AS,\text{ }BP\text{ }=\text{ }BQ,\text{ }CQ\text{ }=\text{ }CR\] and \[DR\text{ }=\text{ }DS\]
\[\therefore \] \[AP\text{ }+\text{ }BP\text{ }+\text{ }CR\text{ }+\text{ }DR\text{ }=\text{ }AS\text{ }+\text{ }BQ\text{ }+\text{ }CQ\text{ }+\text{ }DS\]
\[\Rightarrow \] \[\left( AP\text{ }+\text{ }BP \right)\text{ }+\text{ }\left( CR\text{ }+\text{ }DR \right)\text{ }=\text{ }\left( AS\text{ }+\text{ }DS \right)\text{ }+\text{ }\left( BQ\text{ }+\text{ }CQ \right)\]
\[\Rightarrow \] \[AB\text{ }+\text{ }CD\text{ }=\text{ }AD\text{ }+\text{ }BC\]
4. AB is a diameter and AC is a chord of the circle given below. If \[\angle \,\mathbf{BAC}\text{ }=\text{ }\mathbf{30}{}^\circ \]and the tangent at C intersects AB produced in D, then,
(a) \[BC\text{ }<\text{ }BD\]
(b) \[BC\text{ }>\text{ }BD\]
(c) \[BC\text{ }=\text{ }BD\]
(d) None of these
(e) None of these
Ans. (c)
Explanation: According to the figure \[\angle \,ACB\text{ }=\text{ }90{}^\circ \]
\[\therefore \] \[\angle \,ABC=60{}^\circ \text{ }[\angle \,ABC=180{}^\circ -(Because\,\,\angle \,180{}^\circ \text{ }-\text{ }(\angle \,BAC\text{ }+\angle \,ACB)=180{}^\circ -\left( 30{}^\circ \text{ }+\text{ }90{}^\circ \right)\text{ }=\text{ }60{}^\circ ]\]
And so, \[\angle \,CBD\text{ }=\text{ }120{}^\circ \]
Also, \[\angle \,BCD\text{ }=\angle \,BAC\text{ }=\text{ }30{}^\circ \] (angles in alternate segments)
\[\angle \,BDC\text{ }=\text{ }180{}^\circ \text{ }-\left( 30{}^\circ \text{ }+\text{ }120{}^\circ \right)\text{ }=\text{ }30{}^\circ \]
Thus, \[\angle \,BCD\text{ }=\angle \,BDC.\text{ }\,So,\text{ }BC\text{ }=\text{ }BD.\]
\[\angle \,\mathbf{BCE}\text{ }=\text{ }\mathbf{50}{}^\circ \] then:
(a) ED is not parallel to AB
(b) \[CD\text{ }=\text{ }CE\]
(c) ED passes through the centre of the circle
(d) None of these
Ans. (b)
Explanation: In the following figure Join ED. Then we get,
\[\angle \,DEC\text{ }=\angle \,ACD\text{ }=\text{ }50{}^\circ \] (angles in alternate segments)
And \[\angle \,EDC\text{ }=\angle \,BCE\text{ }=\text{ }50{}^\circ \] (angles in alternate segments)
\[\angle \,DEC\text{ }=\angle \,EDC.\]
So, \[CD\text{ }=\text{ }CE\]
You need to login to perform this action.
You will be redirected in
3 sec