(v) ${{90}^{\circ }}$
Fill in the blanks: (v) The angle between tangent at a point P on circle and radius through the point is _______
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Circles
Fill in the blanks: (iii) A tangent to a circle intersects it in ______ point. (iv) A line intersecting a circle in two points is called a _______
(iii) one point- (iv) secant-
Fill in the blanks: (i) The common point of tangent and the circle is called _________. (ii) A circle may have _____ parallel tangents.
(i) point of contact- (ii) two-
How many tangents can a circle have?
A tangent is characterized as a line crossing the circle in one point. Since, there are limitless number of focuses on the circle, a circle can have many (boundless) digressions.
A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC.
Given: In a circle with centre O, two tangents are drawn from an external point A. An arctangent BC is drawn at a point R with a radius of the circle is 5 cm. Required to find : Perimeter of ∆ABC....
Two circles touch externally at a point P. From a point T on the tangent at P, tangents TQ and TR are drawn to the circles with points of contact Q and R respectively. Prove that TQ = TR.
Given: O and C are the centre of Two circles touching each other externally at P. PT is its common tangent From a point T: PT, TR and TQ are the tangents drawn to the circles. Required to prove: TQ...
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
Given: With OP Joining TS, the circle has a centre O at point P and a tangent to TS. to prove: OP is perpendicular to TS passing through the centre of the circle Construction: Draw a line OR...
In the figure, a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm, and CD = 4 cm. Find AD.
Given, A circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at points P, Q, R and S respectively. $AB=6cm$, $BC=7cm$, $CD=4cm$ Let $AD=X$ As$AP$ and $AS$ are the tangents. $AP=AS$...
AB is a diameter and AC is a chord of a circle with centre O such that ∠BAC = 30°. The tangent at C intersects AB at a point D. Prove that BC = BD.
Required to prove: $BC=BD$ Join $BC$ and$OC$ Given,$\angle BAC={{30}^{\circ }}$ $\Rightarrow \angle BCD={{30}^{\circ }}$ [angle between tangent and chord is equal to angle made by chord in the...
If ∆ABC is isosceles with AB = AC and C (0, r) is the incircle of the ∆ABC touching BC at L. Prove that L bisects BC.
Given: In ∆ABC, AB = AC and O is the centre of the circle and radius (r) touches the side BC of ∆ABC at L. Given to prove : BC’s mid-point is L. Proof : $AM$ and $AN$ are the tangents. So,AN$AM=AN$...
Two tangents segments PA and PB are drawn to a circle with centre O such that ∠APB = 120°. Prove that OP = 2 AP.
Given: PA and PB are tangents and ∠APB = 120° And, OP is joined. Required to prove: OP = 2 AP Construction: Taken M as the mid point of OP and joined AM, join also OA and OB. Proof: In right...
From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP = diameter of the circle, show that ∆APB is equilateral.
Given: O is the centre From a point P outside the circle, PA and PB are the tangents to the circle such that OP is diameter of the circle. And, AB is joined. Given to prove: APB=equilateral triangle...
Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.
Let $M$ is the mid-point of an arc $AMB$ and the tangent to the circle is $TMT'$. Now, join $AB,AM$ and$MB$ Since, arc $AM$ = arc $MB$ ⇒ Chord $AM$ = Chord $MB$ In $\vartriangle AMB,AM=MB$...
In the figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle.
Given, In right $\vartriangle ABC,\angle B={{90}^{\circ }}$ And, $BC=6cm,AB=8cm$ Let us consider, r be the radius of incircle with centre O and touches the sides AB, BC and CA at P, Q and R...
From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of ∆PCD.
Given, PA and PB are the tangents drawn from some extent P outside the circle with centre O. CD is another tangent to the circle at point E, which intersects PA and PB at C and D.. $PA=14cm$ from...
In a right triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that the tangent to the circle at P bisects BC
Center of the given circle is O. At Q,the tangent at P meets. Then join BP. Required to prove: $BQ=QC$ Proof : $\angle ABC={{90}^{\circ }}$ In $\vartriangle ABC,\angle 1+\angle 5={{90}^{\circ }}$...
Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at centre.
We are considering a circle with centre ‘O’ with two parallel tangents through A & B at ends of diameter. M intersect the parallel tangents at P and Q Then, required to prove: $\angle...
If from any point on the common chord of two intersecting circles, tangents be drawn to the circles, prove that they are equal.
Let $X$ and $Y$ are the points at which the two circles intersect. So, $XY$ is the common chord. We might consider $A$ to be a point on the common chord, and $AM$ and $AN$...
A point P is 26 cm away from O of circle and the length PT of the tangent drawn from P to the circle is 10 cm. Find the radius of the circle.
Given, OP = $26cm$ PT = tangent length = $10cm$ To find: radius = OT =$?$ We know that, Radius and tangent are perpendicular at the point of contact, $\angle OTP={{90}^{\circ }}$ $$ So,...
If PT is a tangent at T to a circle whose centre is O and OP = 17 cm, OT = 8 cm. Find the length of the tangent segment PT.
Given in the question, OT = radius = $8cm$ OP = $17cm$ It is given to find: PT = length of tangent =$?$ T is point of contact. We also know that the tangent and radius are perpendicular at the point...