Description : If a given line is tangent to the circle then perpendicular distance from centre of the circle is equal to radius of the circle solve the following :
Last Answer : If a given line is tangent to the circle then perpendicular distance from centre of the circle is equal to radius of the ... 0` D. `x^(2)+y^(2)-8y=0`
Description : If S, L and R are the arc length, long chord and radius of the sliding circle then the perpendicular distance of the line of the resultant cohesive force, is given by (A) a = S.R/L (B) a = L.S/R (C) a = L.R/S (D) None of these
Last Answer : (A) a = S.R/L
Description : What is the distance from a point on the x axis to the centre of a circle when a tangent line at the point 3 4 meets the circle of x2 plus y2 -2x -6y plus 5 equals 0?
Last Answer : Circle equation: x^2 +y^2 -2x -6y +5 = 0Completing the squares: (x-1)^2 +(y-3)^2 = 5Centre of circle: (1, 3)Tangent line meets the x-axis at: (0, 5)Distance from (0, 5) to (1, 3) = 5 units using the distanceformula
Description : Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th
Last Answer : STEP1: Draw a circle with centre at point O and radius 5 cm. STEP2: Draw its cord AB. STEP3: With centre A as centre and radius more than half of AB, draw two arcs, one on each side ... is perpendicular bisector of AB which is chord of circle, Hence, it passes through the centre of the circle.
Description : When a mass is rotating in a plane about a fixed point, its angular momentum is directed along: (1) the tangent to the orbit (2) a line perpendicular to the plane of rotation (3) the line making an angle of 45° to the plane of rotation (4) the radius
Last Answer : (2) a line perpendicular to the plane of rotation
Description : The radius of a circle is 10cm. The length of a chord is 12 cm. Then the distance of the chord from the centre is `"__________________"`.
Last Answer : The radius of a circle is 10cm. The length of a chord is 12 cm. Then the distance of the chord from the centre is `"__________________"`.
Description : A chord of a circle of radius 7.5 cm with centre 0 is of length 9 cm. Find its distance from the centre. -Maths 9th
Last Answer : ∵ PM = MQ = 1/2 = PQ = 45 cm and OP = 7.5 cm In right angled ΔOMP, using phthagoras theorem OM2 = OP2 - PM2 ⇒OM2 = 7.52 - 4.52 ⇒OM2 = 56.25 - 20.25 ⇒OM2 = 36 ∴ OM = √36 = 6 cm
Description : Find the length of a chord which is at a distance of 12 cm from the centre of a circle of radius 13 cm. -Maths 9th
Last Answer : Let AB be a chord of circle with centre O and radius 13cm. Draw OM perpendicular AB and join OA. In the right triangle OMA, we have OA2 = OM2 + AM2 ⇒ 132 = 122 + AM2 ⇒ AM2 = 169 - 144 ... . As the perpendicular from the centre of a chord bisects the chord.Therefore, AB = 2AM = 2 x 5 = 10cm.
Description : The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centre. -Maths 9th
Last Answer : Let PQ be a chord of a circle with centre O and radius 13cm such that PQ = 24cm. From O, draw OM perpendicular PQ and join OP. As, the perpendicular from the centre of a circle to a chord bisects the chord. ∴ PM ... Hence, the distance of the chord from the centre is 5cm.
Description : Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the A distance between AB and CD is 6 cm, find the radius of the circle. -Maths 9th
Last Answer : Join OA and OC. Let the radius of the circle be r cm and O be the centre Draw OP⊥AB and OQ⊥CD. We know, OQ⊥CD, OP⊥AB and AB∥CD. Therefore, points P,O and Q are collinear. So, PQ=6 cm. Let OP=x. Then, ... r2=52+(2.5)2=25+6.25=31.25 ⇒r2=31.25⇒r=5.6 Hence, the radius of the circle is 5.6 cm
Description : The given figure shows a circle with centre O in which a diameter AB bisects the chord PQ at the point R. If PR = RQ = 8 cm and RB = 4 cm, then find the radius of the circle. -Maths 9th
Last Answer : Let r be the radius, then OQ = OB = r and OR = (r - 4) ∴ OQ2 = OR2 + RO2 ⇒ r2 = 64 + (r-4)2 ⇒ r2 = 64 + r2 + 16 - 8r ⇒ 8r = 80 ⇒ r = 10 cm
Description : If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th
Last Answer : According to question the radius of the circle passing through the points A, B and C .
Description : Perpendicular offset from a tangent to the junction of a transition curve and circular curve is equal to (A) Shift (B) Twice the shift (C) Thrice the shift (D) Four times the shift
Last Answer : (D) Four times the shift
Description : If `16l^2+9m^2=24lm+6l+8m+1` and the line `mx+ly=1` is tangent to circle `S_1=0` then the equation of circle `S_2=0` which touches `S_1=0` at `(0,0)`
Last Answer : If `16l^2+9m^2=24lm+6l+8m+1` and the line `mx+ly=1` is tangent to circle `S_1=0` then the equation of ... `3x^(2)+3y^(2)-8x-6y=0` D. none of theese
Description : In the given figure, O is the centre of the circle. The radius OP bisects a rectangle ABCD at right angles. -Maths 9th
Last Answer : answer:
Description : In a circle of radius 14 cm, an arc subtends an angle of 45 O at the centre, then the area of the sector is (a) 71 cm 2(b) 76 cm 2 (c) 77 cm 2 (d) 154 cm 2
Last Answer : (c) 77 cm 2
Last Answer : I determined and what next? A circle inscribed in a triangle This is a circle that touches all sides of the triangle. The center of the circle inscribed in the triangle ABC is the intersection of the axes of the ... the 2nd series of the summer part of KMS 2009 / 2010.pdf example no.6" in Slovak
Description : If is the length of a sub-chord and is the radius of simple curve, the angle of deflection between its tangent and sub-chord, in minutes, is equal to (A) 573 S/R (B) 573 R/S (C) 1718.9 R/S (D) 1718.9 S/R
Last Answer : (D) 1718.9 S/R
Description : If the long chord and tangent length of a circular curve of radius R are equal the angle of deflection, is (A) 30° (B) 60° (C) 90° (D) 120°
Last Answer : D
Description : What is the relationship between chord of a circle and a perpendicular to it from the centre? -Maths 9th
Last Answer : Solution :- Perpendicular line from the centre bisect the chord.
Description : PS is the chord of the circle with centre O. A perpendicular is drawn from centre O of the circle to chord PS at M. If `bar(PS) =30 cm` and `bar(OM) =
Last Answer : PS is the chord of the circle with centre O. A perpendicular is drawn from centre O of the circle to ... =8 cm`, then find the radius of the circle.
Description : What is the length of a tangent line from the point 7 -2 to a point when it touches the circle x2 plus y2 -10 equals 49?
Last Answer : x² + y² - 10 = 49→ x² + y² = 59= (x - 0)² + (y - 0)² = (√59)²→ circle has centre (0, 0) - the origin - and radius √59The point (7, -2) has a distance from the centre of the circleof:√((7 - 0)² + (-2 - 0)²) = √(7² + (-2)²) = √(49 + 4) = √53
Description : What is the length of the tangent line from the point 9 0 to the circle x2 plus 8x plus y2 -9 equals 0?
Last Answer : Equation of circle: x^2 +8x +y^2 -9 = 0Completing the square: (x+4)^2 +y^2 = 25Radius of circle: 5Center of circle: (-4, 0)Distance from (9, 0) to (-4, 0) = 13 which is ... the length of the tangent line is 12 unitsNote that the tangent line of a circle meets the radius of thecircle at right angles
Description : What is the tangent line equation of the circle 2x2 plus 2y2 -8x -5y -1 equals 0 at the point 1 -1 on its circumference?
Last Answer : Circle equation: 2x^2 +2y^2 -8x -5y -1 = 0Divide all terms by 2: x^2 +y^2 -4x -2.5y -0.5 = 0Completing the squares: (x-2)^2 +(y-1.25)^2 = 6.0625Center of circle: (2, 1.25)Point of contact: (1 ... 9Tangent equation: y--1 = -4/9(x-1) => 9y = -4x -5Tangent line equation in its general form: 4x+9y+5 = 0
Description : What are the values of x y and k when the line y equals 3x plus 1 is a tangent to the circle x squared plus y squared equals k?
Last Answer : If y = 3x + 1 is a tangent to x² + y² = k (k > 0 since it isa square), then where they meet has a repeated root; they meetat:x² + (3x + 1)² = k→ x² + 9x² + 6x + 1 - k = 0→ 10x² + 6x + (1 - k) = 0This is the ... 3/10→ y = 3 -3/10 + 1 = 9/10 + 1 = 1/10→ point of contact is (-3/10, 1/10) with k = 1/10
Description : What is the equation of the tangent line that meets the circle x2 -4x plus y2 -6y equals 4 at the point 6 4?
Last Answer : Circle equation: x^2 -4x +y^2 -6y = 4Completing the squares: (x-2)^2 +(y-3)^2 = 17Point of contact: (6, 4)Center of circle: (2, 3)Slope of radius: 1/4Slope of tangent line: -4Tangent equation: y-4 = -4(x-6) => y = -4x+28Tangent line equation in its general form: 4x+y-28 = 0
Description : What is the length of the tangent line from the point 8 2 to a point where it meets the circle x2 plus y2 -4x -8y -5 equals 0?
Last Answer : Equation of circle: x^2 +y^2 -4x -8y -5 = 0Completing the squares: (x-2)^2 +(y-4)^2 = 25 which is radiussquaredCenter of circle: (2, 4)Tangent line originates from: (8, ... angle triangleUsing Pythagoras theorem: distance^2 minus radius^2 = 15Therefore length of tangent line is the square root of 15
Description : What is the point of contact when the tangent line y equals x plus c touches the circle x squared plus y squared equals 4?
Last Answer : The line meets the circle when:y = x + c→ x² + y² = 4→ x² + (x + c)² - 4 = 0→ x² + x² +2cx + c² - 4 = 0→ 2x² + 2cx + (c² - 4) = 0If the line is a tangent to the circle, this has a repeatedroot. This occurs ... 2 = 0→ (x + √2)² = 0→ x = -√2→ y = x + 2√2= -√2 + 2√2= √2→ point of contact is (-√2, √2)
Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th
Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.
Description : A sphere of radius r is cut from larger sphere of radius R. The distance between their centre is a. The centroid of the remaining valoume lies on the line of centres and the sistanc from the centre of the larger sphere is a.ar2 / (R2 - r2) b.107 dynes c.aR/(R2-r2) d.aR/(R - r) e.ar3 / (R3 - r3)
Last Answer : e. ar3 / (R3 - r3)
Description : The slope of one of the common tangent to circle `x^(2)+y^(2)=1` and ellipse `x^(2)/4+2y^(2)=1` is `sqrt(a/b)` where `gcd(a, b)=1` then `(a+b)/2` is e
Last Answer : The slope of one of the common tangent to circle `x^(2)+y^(2)=1` and ellipse `x^(2)/4+2y^(2)=1` is ` ... `gcd(a, b)=1` then `(a+b)/2` is equal to
Description : Find the co-ordinates of that point on the curve`y^(2)=x^(2)(1-x)` at which the tangent drawn is perpendicular to X-axis.
Last Answer : Find the co-ordinates of that point on the curve`y^(2)=x^(2)(1-x)` at which the tangent drawn is perpendicular to X-axis.
Description : The approximate formula for radial or perpendicular offsets from the tangent, is (A) x/2R (B) x²/2R (C) x/R (D) x²/R
Last Answer : (B) x²/2R
Description : If a gauge is installed perpendicular to the slope, its measurement is reduced by multiplying (A) Sine of the angle of inclination with vertical (B) Cosine of the angle of inclination with vertical (C) Tangent of the angle of inclination with vertical (D) Calibration coefficient of the gauge
Last Answer : Answer: Option B
Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th
Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm
Description : Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle. -Maths 9th
Last Answer : clear .
Description : AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre. Prove that -Maths 9th
Last Answer : Draw OM perpendicular AB and ON perpendicular AC Join OA. In right △OAM, OA2 = OM2 + AM2 ⇒ r2 = p2 + (1/2AB)2 (Since,OM perpendicular AB, ∴ OM bisects AB ) ⇒ 1/4AB2 = r2 - p2 or AB2 = 4r2 - 4p2 ... ) and (ii), we have 4r2 - 4p2 = 16r2 - 16q2 or r2 - p2 = 4r2 - 4q2 or 4q2 = 3r2 + p2
Description : Point A(5, 1) is the centre of the circle with radius 13 units. AB ⊥ chord PQ. B is (2, –3). The length of chord PQ is -Maths 9th
Last Answer : (c) ParallelogramAB = \(\sqrt{(4-7)^2+(5-6)^2}\) = \(\sqrt{9+1}\) = \(\sqrt{10}\)BC = \(\sqrt{(7-4)^2+(6-3)^2}\) = \(\sqrt{9+9}\) = \(3\sqrt2\)CD =\(\sqrt{(4-1)^2+(3-2) ... = \(2\sqrt{13}\)AB = CD, BC = AD and AC ≠ BD ⇒ opposite sides are equal and diagonals are not equal. ⇒ ABCD is a parallelogram.
Description : A chord of a circle of radius 20 cm subtends an angle of 90° at the centre. Find the area of the corresponding major segment of the circle. -Maths 10th
Last Answer : Area of the minor segment = { pi × 90 /360 - sin 45 × cos 45 } × r × r ={ 3.14 ×1/4 - 1÷√2 ×1÷√2 } × 20 × 20 = { 3.14 ... Area of major segment = area of circle - area of minor segment = 1256 - 114 = 1142 HOPE IT HELPS YOU
Description : In the figure below, CD is a chord of the semi circle with centre O. OA is the radius of the circle. If `CD=10` cm, `AB=2` cm and `bar(OA)_|_bar(CD)`
Last Answer : In the figure below, CD is a chord of the semi circle with centre O. OA is the radius of the ... |_bar(CD)` the length of OB is `"_____________"`
Description : An electron moves in a circular orbit with a uniform speed `v`.It produces a magnetic field `B` at the centre of the circle. The radius of the circle
Last Answer : An electron moves in a circular orbit with a uniform speed `v`.It produces a magnetic field `B` at the centre of the ... (v)/(B))` D. `sqrt((B)/(v))`
Description : In the following figure a wire bent in the form of a regular polygon of `n` sides is inscribed in a circle of radius `a`. Net magnetic field at centre
Last Answer : In the following figure a wire bent in the form of a regular polygon of `n` sides is inscribed in a circle of ... (ni)/(a)mu_(0)tan.(pi)/(n)` D.
Description : What is the radius of the centre circle in football?
Last Answer : Need answer
Description : To set out a parallel from a given inaccessible point to a given line AB, the following observations are made: Distance AB and angle PAM = a and angle PBA = b are measured where M is a point on the line BA produced. The ... cos b - cos a) (C) AB/(cot a - cot b) (D) AB/(cot a - cos b)
Last Answer : (A) AB/(cot b - cot a)
Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th
Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.