Mix Examples - Circles Questions in English

(D) According to the theorem of circles,the tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,the angle formed between the tangent and the radius at the point of contact is $90^{\circ}$.

(C) According to the theorem of tangents drawn from an external point to a circle,the lengths of tangents drawn from an external point to a circle are equal.
Since $P$ is an external point and $\overline{ PA }$ and $\overline{ PB }$ are tangents to the circle $\odot( O , r)$,it follows that $PA = PB$.

(A) Let $O$ be the center of the circle. Since $AB$ is the diameter,$OA$ and $OB$ are radii of the circle.
By the property of tangents,a tangent at any point of a circle is perpendicular to the radius through the point of contact.
Therefore,$l_{1} \perp OA$ and $l_{2} \perp OB$.
Since $A, O,$ and $B$ are collinear (as $AB$ is a diameter),the lines $OA$ and $OB$ form a single straight line $AB$.
If two lines ($l_{1}$ and $l_{2}$) are perpendicular to the same line $(AB)$,then they must be parallel to each other.
Thus,$l_{1} \parallel l_{2}$.

(C) The radius of the circle is $PB = 5$. The tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore,$\angle PBA = 90^\circ$.
In the right-angled triangle $\triangle PBA$,by the Pythagoras theorem:
$PA^2 = PB^2 + AB^2$
$13^2 = 5^2 + AB^2$
$169 = 25 + AB^2$
$AB^2 = 169 - 25 = 144$
$AB = \sqrt{144} = 12$.

(B) According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.
Therefore,$PA = PB$.
Given that $PA = 8$,it follows that $PB = 8$.

(C) Given that the radius of the circle $OM = 8$ and $NP = 2$.
Since $ON$ is the radius,$ON = 8$.
Therefore,the distance $OP = ON + NP = 8 + 2 = 10$.
In the right-angled triangle $\Delta OMP$,the angle $\angle OMP = 90^{\circ}$ because the tangent is perpendicular to the radius at the point of contact.
Using the Pythagorean theorem: $OP^2 = OM^2 + PM^2$.
$10^2 = 8^2 + PM^2$.
$100 = 64 + PM^2$.
$PM^2 = 100 - 64 = 36$.
$PM = \sqrt{36} = 6$.

(D) In the quadrilateral $OAPB$,the radius is perpendicular to the tangent at the point of contact.
Therefore,$\angle OAP = 90^{\circ}$ and $\angle OBP = 90^{\circ}$.
The sum of the angles in a quadrilateral is $360^{\circ}$.
Thus,$\angle AOB + \angle OAP + \angle APB + \angle OBP = 360^{\circ}$.
Substituting the known values: $\angle AOB + 90^{\circ} + 65^{\circ} + 90^{\circ} = 360^{\circ}$.
$\angle AOB + 245^{\circ} = 360^{\circ}$.
$\angle AOB = 360^{\circ} - 245^{\circ} = 115^{\circ}$.

(D) The line segment $OP$ bisects the angle $\angle APB$.
Therefore,$m \angle OPB = \frac{1}{2} m \angle APB = \frac{1}{2} (70^\circ) = 35^\circ$.
Since $PB$ is a tangent to the circle at point $B$,the radius $OB$ is perpendicular to the tangent $PB$.
Thus,$m \angle OBP = 90^\circ$.
In the right-angled triangle $\Delta OBP$,the sum of the angles is $180^\circ$.
$m \angle POB + m \angle OBP + m \angle OPB = 180^\circ$.
$m \angle POB + 90^\circ + 35^\circ = 180^\circ$.
$m \angle POB + 125^\circ = 180^\circ$.
$m \angle POB = 180^\circ - 125^\circ = 55^\circ$.

(B) In $\Delta OAP$,the radius $OA$ is perpendicular to the tangent $PA$ at the point of contact $A$.
Therefore,$m\angle OAP = 90^\circ$.
In $\Delta OAP$,the sum of the angles is $180^\circ$.
$m\angle OAP + m\angle AOP + m\angle OPA = 180^\circ$
$90^\circ + 40^\circ + m\angle OPA = 180^\circ$
$130^\circ + m\angle OPA = 180^\circ$
$m\angle OPA = 180^\circ - 130^\circ = 50^\circ$

(D) In $\Delta OAP$,the radius $OA$ is perpendicular to the tangent $PA$ at the point of contact $A$. Therefore,$m\angle OAP = 90^\circ$.
Given $m\angle AOP = 45^\circ$.
In $\Delta OAP$,the sum of angles is $180^\circ$.
$m\angle OPA = 180^\circ - (90^\circ + 45^\circ) = 180^\circ - 135^\circ = 45^\circ$.
Since $m\angle AOP = m\angle OPA = 45^\circ$,$\Delta OAP$ is an isosceles triangle.
Therefore,the sides opposite to equal angles are equal: $AP = OA$.
Since $OA$ is the radius of the circle,$OA = 8$.
Thus,$AP = 8$.

(C) Since the tangent at any point of a circle is perpendicular to the radius through the point of contact,$\angle OAP = 90^{\circ}$.
In right-angled triangle $\triangle OAP$,by Pythagoras theorem:
$OP^{2} = OA^{2} + AP^{2}$
Given $OP = 10$ and $AP = 8$,we have:
$10^{2} = OA^{2} + 8^{2}$
$100 = OA^{2} + 64$
$OA^{2} = 100 - 64 = 36$
$OA = \sqrt{36} = 6$
Here,$OA$ is the radius $(r)$ of the circle.
Diameter of the circle $= 2 \times \text{radius} = 2 \times 6 = 12$.

(C) Let $O$ be the common center of the two concentric circles.
Let $AB$ be the chord of the larger circle which touches the smaller circle at point $M$.
$OM$ is the radius of the smaller circle,so $OM = 5$.
$OB$ is the radius of the larger circle,so $OB = 13$.
Since the radius is perpendicular to the tangent at the point of contact,$\angle OMB = 90^{\circ}$.
In right-angled $\Delta OMB$,by Pythagoras theorem:
$OB^2 = OM^2 + MB^2$
$13^2 = 5^2 + MB^2$
$169 = 25 + MB^2$
$MB^2 = 169 - 25 = 144$
$MB = \sqrt{144} = 12$.
The perpendicular from the center to a chord bisects the chord.
Therefore,$AB = 2 \times MB = 2 \times 12 = 24$.

(D) The sides of $\Delta ABC$ are $a=5$,$b=12$,and $c=13$.
Since $5^2 + 12^2 = 25 + 144 = 169 = 13^2$,$\Delta ABC$ is a right-angled triangle where the hypotenuse is $13$.
The radius $r$ of the incircle of a right-angled triangle is given by the formula $r = \frac{a+b-c}{2}$,where $a$ and $b$ are the legs and $c$ is the hypotenuse.
Substituting the values: $r = \frac{5+12-13}{2} = \frac{17-13}{2} = \frac{4}{2} = 2$.
Thus,the radius of the circle touching all three sides is $2$.

(D) In $\Delta PQR$,$\angle Q = 90^{\circ}$.
Therefore,$\overline{PR}$ is the hypotenuse.
By Pythagoras theorem,$PR^2 = PQ^2 + QR^2$.
$PR^2 = 8^2 + 15^2 = 64 + 225 = 289$.
$PR = \sqrt{289} = 17$.
The radius $(r)$ of the incircle of a right-angled triangle is given by the formula $r = \frac{PQ + QR - PR}{2}$.
$r = \frac{8 + 15 - 17}{2} = \frac{23 - 17}{2} = \frac{6}{2} = 3$.

(B) tangent to a circle is a line that intersects the circle at exactly one point.
From any point $P$ lying in the exterior of a circle,exactly two tangents can be drawn to the circle.
These two tangents touch the circle at two distinct points.
Therefore,the maximum number of tangents that can be drawn from an external point to a circle is $2$.

(C) Let the lengths of the tangents from vertices $A, B, C$ to the circle be $x, y, z$ respectively.
Since tangents drawn from an external point to a circle are equal in length,we have:
$AD = AF = x$
$BD = BE = y$
$CE = CF = z$
Given the side lengths of $\Delta ABC$:
$AB = x + y = 13$
$BC = y + z = 12$
$CA = z + x = 5$
Adding these three equations:
$(x + y) + (y + z) + (z + x) = 13 + 12 + 5$
$2(x + y + z) = 30$
$x + y + z = 15$
We need to find $AD = x$. Using $y + z = 12$:
$x + (y + z) = 15$
$x + 12 = 15$
$x = 15 - 12 = 3$
Therefore,$AD = 3$.

(D) The circle $\odot(O, 5)$ has a radius $r = 5$.
Since the circle touches all sides of the square,the diameter of the circle is equal to the side length of the square.
Side length of the square $= 2 \times r = 2 \times 5 = 10$.
The perimeter of a square $= 4 \times \text{side length} = 4 \times 10 = 40$.

(D) quadrilateral that circumscribes a circle is known as a tangential quadrilateral. According to the Pitot theorem,for a quadrilateral $ABCD$ that touches a circle on all four sides,the sums of the lengths of opposite sides are equal. That is,$AB + CD = BC + DA$. Since the question asks for the classification of such a quadrilateral,the most general term is a tangential quadrilateral. If the options provided are limited,note that while a square and a rhombus are specific types of tangential quadrilaterals,the property of touching all sides defines a tangential quadrilateral.

(C) cyclic quadrilateral that is also a rectangle implies that its vertices lie on a circle and its interior angles are $90^{\circ}$.
In a rectangle $ABCD$,$\Delta ABC$ is a right-angled triangle with $\angle B = 90^{\circ}$.
According to the Pythagorean theorem,$AC^{2} = AB^{2} + BC^{2}$.
Substituting the given values: $AC^{2} = 5^{2} + 12^{2}$.
$AC^{2} = 25 + 144 = 169$.
Therefore,$AC = \sqrt{169} = 13$.

(D) Let $O$ be the common center of the two concentric circles. The radius of the larger circle is $OB = 41$ and the radius of the smaller circle is $OM = 9$.
Since the chord $\overline{AB}$ of the larger circle is tangent to the smaller circle at $M$,the radius $OM$ is perpendicular to the chord $AB$ at the point of tangency $M$.
Thus,in the right-angled triangle $\Delta OMB$,by the Pythagorean theorem:
$OB^2 = OM^2 + MB^2$
$41^2 = 9^2 + MB^2$
$1681 = 81 + MB^2$
$MB^2 = 1681 - 81 = 1600$
$MB = \sqrt{1600} = 40$.
Since the perpendicular from the center to a chord bisects the chord,$AB = 2 \times MB = 2 \times 40 = 80$.

(C) Let $O$ be the common center of the two concentric circles.
Let $R = 17$ be the radius of the outer circle and $r = 15$ be the radius of the inner circle.
The chord $\overline{AB}$ of the outer circle is tangent to the inner circle at point $M$.
Since the radius is perpendicular to the tangent at the point of contact,$OM \perp AB$.
In $\triangle OMA$,by the Pythagorean theorem: $OA^2 = OM^2 + AM^2$.
Substituting the values: $17^2 = 15^2 + AM^2$.
$289 = 225 + AM^2$.
$AM^2 = 289 - 225 = 64$.
$AM = \sqrt{64} = 8$.
Since the perpendicular from the center to a chord bisects the chord,$AB = 2 \times AM$.
$AB = 2 \times 8 = 16$.

(B) circle touches all the four sides of $\square ABCD$. According to the property of a tangential quadrilateral,the sums of opposite sides are equal.
$\therefore AB + CD = AD + BC$
Substituting the given values:
$\therefore 5 + 6 = AD + 8$
$\therefore 11 = AD + 8$
$\therefore AD = 11 - 8 = 3$

(C) $\square PQRS$ is a cyclic quadrilateral.
In a cyclic quadrilateral, the sum of the measures of opposite angles is $180^{\circ}$.
Therefore, $m \angle P + m \angle R = 180^{\circ}$.
Given that $m \angle P = 30^{\circ}$.
Substituting the value, we get $30^{\circ} + m \angle R = 180^{\circ}$.
Thus, $m \angle R = 180^{\circ} - 30^{\circ} = 150^{\circ}$.

(A) Let the tangent $\overleftrightarrow{PQ}$ touch the circle at point $R$.
Since tangents drawn from an external point to a circle are equal in length,we have:
$AB = AC$ (tangents from point $A$)
$PB = PR$ (tangents from point $P$)
$QC = QR$ (tangents from point $Q$)
The perimeter of $\Delta APQ = AP + PQ + AQ$.
Substituting $PQ = PR + QR$,we get:
Perimeter $= AP + (PR + QR) + AQ$.
Using the equalities $PR = PB$ and $QR = QC$:
Perimeter $= AP + PB + QC + AQ$.
Since $AP + PB = AB$ and $AQ + QC = AC$:
Perimeter $= AB + AC$.
Since $AB = AC$,the perimeter $= AB + AB = 2 AB$ (or $2 AC$).
Thus,the correct option is $2 AB$.

(A) For a quadrilateral circumscribing a circle,the sum of opposite sides is equal,i.e.,$AB + CD = BC + AD$.
Given that $\overline{AB}$ is the largest side,let $AB = x$.
Since the circle is tangent to all sides,the lengths of the tangents from the vertices to the circle are equal. Let the lengths of the tangents from vertices $A, B, C, D$ be $a, b, c, d$ respectively.
Then $AB = a + b$,$BC = b + c$,$CD = c + d$,and $AD = a + d$.
Since $AB$ is the largest side,$a+b$ is the maximum value.
In a tangential quadrilateral,the side opposite to the largest side is generally the smallest side. Thus,$\overline{CD}$ is the smallest side.

(C) Let the tangent $\overleftrightarrow{PQ}$ touch the circle at point $R$.
Since the lengths of tangents drawn from an external point to a circle are equal,we have:
$PB = PR$ and $QC = QR$.
The perimeter of $\Delta APQ = AP + PQ + AQ$.
$= AP + (PR + RQ) + AQ$.
$= AP + PB + QC + AQ$ (since $PR = PB$ and $RQ = QC$).
$= (AP + PB) + (AQ + QC)$.
$= AB + AC$.
Since $AB = AC$ (tangents from an external point $A$),we have $AB = AC = 6$.
Therefore,the perimeter of $\Delta APQ = 6 + 6 = 12$.

(C) Given that $\overline{ PA }$ and $\overline{ PB }$ are tangents to the circle $\odot( O , r)$ from an external point $P$.
According to the theorem,the lengths of tangents drawn from an external point to a circle are equal.
Therefore,$\overline{ PA } = \overline{ PB }$.
In $\Delta PAB$,since $\overline{ PA } = \overline{ PB }$,the triangle is an isosceles triangle.
In an isosceles triangle,the angles opposite to the equal sides are equal.
Thus,$m \angle PAB = m \angle PBA$.
Given that $m \angle PAB = 60^{\circ}$,it follows that $m \angle PBA = 60^{\circ}$.

(C) Let the tangents from point $A$ to the circle be $AB$ and $AC$. By the theorem,lengths of tangents drawn from an external point to a circle are equal,so $AB = AC$.
Let the tangent $PQ$ touch the circle at point $M$. Then $PM = PB$ and $QM = QC$ (tangents from points $P$ and $Q$ respectively).
The perimeter of $\Delta APQ = AP + PQ + AQ = AP + (PM + MQ) + AQ$.
Substituting the equal tangent lengths: Perimeter $= AP + PB + QC + AQ = (AP + PB) + (AQ + QC) = AB + AC$.
Since $AB = AC$,the perimeter $= 2 AB$.
Given that the perimeter is $16$,we have $2 AB = 16$,which implies $AB = 8$.

(C) In the given circle,$\overline{ OB }$ is the radius and $\overline{ PB }$ is the tangent at point $B$.
Since the radius is perpendicular to the tangent at the point of contact,$\angle OBP = 90^{\circ}$.
Thus,$\Delta POB$ is a right-angled triangle with hypotenuse $OP$.
By the Pythagorean theorem:
$OP^{2} = OB^{2} + PB^{2}$
$13^{2} = 5^{2} + PB^{2}$
$169 = 25 + PB^{2}$
$PB^{2} = 169 - 25 = 144$
$PB = \sqrt{144} = 12$
Therefore,$PB = 12$.

(B) When two circles touch each other externally,they have one direct common tangent passing through the point of contact and two transverse common tangents.
Therefore,a total of $3$ common tangents can be drawn to the two circles.

(C) tangent to a circle is perpendicular to the radius at the point of contact.
Here,$\stackrel{\leftrightarrow}{AB}$ is a tangent to the circle $\odot(P, r)$ at point $Q$.
Therefore,the radius $PQ$ is perpendicular to the tangent $\stackrel{\leftrightarrow}{AB}$ at $Q$.
Since $PQ \perp AB$,the foot of the perpendicular drawn from $P$ to the line $AB$ is $Q$.

(C) The distance from the centre $P$ of the circle to the line $AB$ is the length of the perpendicular segment $PQ$.
Given that $Q$ lies in the interior of the circle,the length of $PQ$ must be less than the radius $r$ of the circle (i.e.,$PQ < r$).
According to the properties of a circle,if the perpendicular distance from the centre to a line is less than the radius of the circle,then the line is a secant line and intersects the circle at two distinct points.

(B) Let the sides of the right-angled triangle be $a = BC$,$c = AB$,and the hypotenuse $b = AC$.
Let the radius of the incircle be $r$.
The center of the incircle $(O)$ forms a square with the vertices at the point of tangency on sides $AB$ and $BC$ and the vertex $B$ because the radii are perpendicular to the sides.
Thus,the distance from $B$ to the points of tangency on $AB$ and $BC$ is $r$.
The remaining segments of the sides $AB$ and $BC$ are $(c - r)$ and $(a - r)$ respectively.
By the property of tangents drawn from an external point to a circle,these segments are equal to the distances from vertices $A$ and $C$ to the points of tangency on the hypotenuse.
Therefore,the hypotenuse $b = (c - r) + (a - r)$.
$b = a + c - 2r$.
$2r = a + c - b$.
$r = \frac{a + c - b}{2}$.
Substituting the side lengths: $r = \frac{BC + AB - AC}{2}$.

(B) In $\Delta ABC$,$m\angle B = 90^{\circ}$,$AB = 24$,and $BC = 7$.
Using the Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (24)^2 + (7)^2$
$AC^2 = 576 + 49$
$AC^2 = 625$
$AC = \sqrt{625} = 25$.
The radius $r$ of the incircle of a right-angled triangle is given by the formula:
$r = \frac{AB + BC - AC}{2}$
$r = \frac{24 + 7 - 25}{2}$
$r = \frac{31 - 25}{2}$
$r = \frac{6}{2} = 3$.
Therefore,the radius of the circle is $3$.

(D) quadrilateral that circumscribes a circle is known as a tangential quadrilateral.
According to the properties of tangents drawn from an external point to a circle, the lengths of tangents are equal.
Let the circle touch the sides $AB$, $BC$, $CD$, and $DA$ at points $P$, $Q$, $R$, and $S$ respectively.
Then, $AP = AS$, $BP = BQ$, $CQ = CR$, and $DR = DS$.
Adding these equations: $(AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)$ implies $AB + CD = AD + BC$.
This property is satisfied by any tangential quadrilateral.
However, among the given options, a rhombus is a specific type of quadrilateral where all sides are equal and it can circumscribe a circle.
While a square is also a rhombus, the most general classification for a quadrilateral whose sides are tangent to a circle is a tangential quadrilateral, but in the context of standard geometry problems of this type, the property $AB + CD = AD + BC$ is the defining characteristic.
Given the options, a rhombus is the most appropriate geometric figure that satisfies the condition of having an incircle.

(B) In a right-angled triangle,the radius $r$ of the incircle is given by the formula $r = \frac{AB + BC - AC}{2}$.
First,we find the hypotenuse $AC$ using the Pythagorean theorem: $AC = \sqrt{AB^2 + BC^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.
Now,substitute the values into the formula: $r = \frac{4 + 3 - 5}{2} = \frac{7 - 5}{2} = \frac{2}{2} = 1$.
Therefore,the radius of the incircle is $1$.

(A) rhombus is a quadrilateral with all sides equal.
If a rhombus is inscribed in a circle,all its vertices lie on the circumference of the circle.
$A$ quadrilateral inscribed in a circle is called a cyclic quadrilateral.
In a cyclic quadrilateral,the sum of opposite angles is $180^{\circ}$.
For a rhombus $ABCD$,the opposite angles are equal,i.e.,$\angle A = \angle C$ and $\angle B = \angle D$.
Since $\angle A + \angle C = 180^{\circ}$,we have $2\angle A = 180^{\circ}$,which implies $\angle A = 90^{\circ}$.
$A$ rhombus with an angle of $90^{\circ}$ is a square.
Therefore,the rhombus $ABCD$ must be a square.

(D) For a right-angled triangle with sides $a, b$ and hypotenuse $c$,the in-radius $r$ is given by the formula $r = \frac{a+b-c}{2}$.
$1.$ For $\Delta ABC$ with sides $3, 4, 5$: $r = \frac{3+4-5}{2} = \frac{2}{2} = 1$. Matches $(b)$.
$2.$ For $\Delta PQR$ with sides $5, 12, 13$: $r = \frac{5+12-13}{2} = \frac{4}{2} = 2$. Matches $(a)$.
$3.$ For $\Delta XYZ$ with sides $8, 15, 17$: $r = \frac{8+15-17}{2} = \frac{6}{2} = 3$. Matches $(d)$.
$4.$ For $\Delta MNP$ with sides $20, 21, 29$: $r = \frac{20+21-29}{2} = \frac{12}{2} = 6$. Matches $(c)$.
Thus,the correct matching is $(1-b), (2-a), (3-d), (4-c)$.