Category : 9th Class
Theorem 1
A tangent of a circle is perpendicular to the radius through the point of contact
Given: I is the tangent to the circle c (o, r) at point A
To prove: \[OA\bot \ell \]
Construction:
Take a point B on line I other than A and join OB. Let OB cut the circle at the point C.
Proof:
Since O does not lies l. Therefore, the shortest distance from 0 to I will be the perpendicular so it is sufficient to prove OA is the shortest distance.
From figure
\[\text{OA}=\text{OC}\] \[\Rightarrow \] \[~\text{OA}<\text{PN}\]
OA is the shorter than any other segment joining 0 to any point on I. Hence, \[~\text{OA}\bot \ell \]
Theorem 2
The length of tangent drawn from an external point to a circle are equal.
Given: PA and PB are the tangents from point P to circle C (o, r).
To Prove: PA = PB
Construction:
Join PO, OA and BO
Proof:
In \[\Delta \text{POA}\] and \[\Delta \text{POB}\]
\[\text{PO}=\text{OP}\] (common)
\[\text{OA}=\text{OB}\] (radius of circle)
\[\angle \text{OAP}=\angle \text{OBP}=\text{9}0{}^\circ \]
By R.H.S Criteria
\[\Delta \text{POA}\cong \Delta \text{POB}\]
\[\frac{\text{By C}\text{.P}.\text{C}\text{.T}}{\text{PA}=\text{PB}}\]
Hence, proved
Another method:
In\[\Delta \text{POA}\]
\[P{{A}^{2}}+O{{A}^{2}}=P{{O}^{2}}\] ...(i)
In \[\Delta \text{POB}\]
\[\text{P}{{\text{B}}^{\text{2}}}+\text{O}{{\text{B}}^{\text{2}}}=\text{P}{{\text{O}}^{\text{2}}}\] ...(ii)
From (i) and (ii) we get
\[\text{P}{{\text{A}}^{\text{2}}}+\text{O}{{\text{A}}^{\text{2}}}=\text{P}{{\text{B}}^{\text{2}}}+\text{O}{{\text{B}}^{\text{2}}}\]
\[\Rightarrow \]\[\text{P}{{\text{A}}^{\text{2}}}=\text{P}{{\text{B}}^{\text{2}}}\](Since\[\text{O}{{\text{A}}^{\text{2}}}=\text{O}{{\text{B}}^{\text{2}}}\], because they are radius) \[\Rightarrow \] \[\text{PA}=\text{PB}\]
Hence, proved.
In an isosceles A PQR in which PQ = QR. A circle which passes through P and R intersect at point L and M to the side PQ and QR respectively as shown in the figure then which one of the following is true?
(a) L and M are the mid points of PQ and QR respectively
(b) \[\text{LM}=\frac{1}{2}\text{PR}\]
(c) \[\text{LM}|\,\,|\text{PR}\]
(d) \[\angle PQR=\angle QLM\]
(e) None of these
Answer: (c)
Explanation:
Since, \[\Delta \text{PQR}\] is isosceles triangle in which \[PQ=QR\] \[\Rightarrow \] \[\angle QPR=\angle QRP\]
Since, LMRP is a cyclic quadrilateral
\[\angle \text{LPR}+\angle \text{LMR}=\text{18}0{}^\circ \] .... (i)
and \[\angle \text{MLP}+\angle \text{MRP}=\text{18}0{}^\circ \] ....(ii)
From (i) and (ii), we get \[\angle \text{MLP}=\angle \text{LMR}\] ...(iii)
and \[\text{QML}+\text{LMR}=\text{l8}0{}^\circ \] ....(iv)
From (i) and (iv)
\[\Rightarrow \]\[\angle \text{QML}+\angle \text{LMR}=\angle \text{LPR}+\angle \text{LMR}\]
\[\Rightarrow \] \[\angle \text{QML}=\angle \text{LPR}\] \[\Rightarrow \] \[\angle \text{QML}=\angle \text{QRP}\] [because \[\angle \text{LPR}=\angle \text{QRP}\]]
Since, corresponding angles are equal
Therefore, \[LM\,|\,\,|\,\,QR\]
In the figure given below PQ is the diametre of the circle with center O. If PT and QS are the perpendicular to the line I, and QS meet the circle at point R then which one of the following is correct?
(a) \[\Delta \text{PRT}\] is acute angled triangle
(b) TP = RS
(c) Quadrilateral RPST is a square
(d) \[\angle \text{SRP}\] is obtuse angle
(e) None of these
Answer: (b)
In the figure given below which one of the following relations is correct?
(a) \[\angle m=\angle l+\angle n\]
(b) \[\angle l=\angle m-\angle n\]
(c) \[\angle m=\angle n-\angle l\]
(d) \[\angle n=\angle m-\angle l\]
(e) None of these
Answer: (c)
Find the radius of circle if a chord of length is 30cm which is drawn at a distance of 8 cm from centre of circle.
(a) 17cm
(b) 15cm
(c) 4cm
(d) 12cm
(e) None of these
Answer: (a)
In the figure given below AP and AQ are tangents from point A to a circle whose centre is O. Another tangents through M intersects AP and AO at L & N respectively and tangents at y intersects at x and z respectively then which one of the options is correct.
(a) XZ=PX+QZ
(b) LN=LM+PL
(c) AL=AN
(d) XZ=QZ+YZ
(d) None of these
Answer: (a)
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