In the adjoining figure, M is the center of the circle and seg KL is a

1.	In the adjoining figure, M is the center of the circle and seg KL is a tangent segment. If MK = 12, KL = 6√3 ,then find i. Radius of the circle. ii. Measures of ∠K and ∠M.
Answer» i. Line KL is the tangent to the circle at point L and seg ML is the radius. [Given] ∴ ∠MLK = 90°…………. (i) [Tangent theorem] In ∆MLK, ∠MLK = 90° ∴ MK² = ML² + KL² [Pythagoras theorem] ∴ 12² = ML² + (6√3)² ∴ 144 = ML² + 108 ∴ ML² = 144 – 108 ∴ ML² = 36 ∴ ML = √36 = 6 units. [Taking square root of both sides] ∴ Radius of the circle is 6 units. ii. We know that, ML = 1/2 MK ∴ ∠K = 30° …………… (ii) [Converse of 30° – 60° – 90° theorem] In ∆MLK , ∠L = 90° [From (i)] ∠K = 30° [From (ii)] ∴ ∠M = 60° [Remaining angle of ∆MLK]

In the adjoining figure, M is the center of the circle and seg KL is a tangent segment. If MK = 12, KL = 6√3 ,then find i. Radius of the circle. ii. Measures of ∠K and ∠M.

Answer»

i. Line KL is the tangent to the circle at point L and seg ML is the radius. [Given]

∴ ∠MLK = 90°…………. (i) [Tangent theorem]

In ∆MLK, ∠MLK = 90°

∴ MK² = ML² + KL² [Pythagoras theorem]

∴ 12² = ML² + (6√3)²

∴ 144 = ML² + 108

∴ ML² = 144 – 108

∴ ML² = 36

∴ ML = √36 = 6 units. [Taking square root of both sides]

∴ Radius of the circle is 6 units.

ii. We know that,

ML = 1/2 MK

∴ ∠K = 30° …………… (ii) [Converse of 30° – 60° – 90° theorem]

In ∆MLK ,

∠L = 90° [From (i)]

∠K = 30° [From (ii)]

∴ ∠M = 60° [Remaining angle of ∆MLK]

Discussion

No Comment Found

Related InterviewSolutions

The circle described on the line joining the points `(0,1), (a, b)` as diameter cuts the x-axis in points whose abscissae are roots of the equationA. `x^(2)+ax+b=0`B. `x^(2)-ax+b=0`C. `x^(2)+ax-b=0`D. `x^(2)-ax-b=0`
The equation of the circle whose one diameter is PQ, where the ordinates of P, Q are the roots of the equation `x^(2)+2x-3=0` and the abscissae are the roots of the equation `y^(2)+4y-12=0` isA. `x^(2)+y^(2)+2x+4y-15=0`B. `x^(2)+y^(2)-4x-2y-15=0`C. `x^(2)+y^(2)+4x+2y-15=0`D. none of these
Statement-1: The equation of a circle through the origin and belonging to the coaxial system, of which limiting points are (1, 1) and (3, 3) is `2x^(2)+2y^(2)-3x-3y=0` Statement-2: The equation of a circle passing through the points (1, 1) and (3, 3) is `2x^(2)+y^(2)-2x-6y+6=0`.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement-2 is not a correct explanation for Statement-1.C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True.
A circle touches y-axis at (0, 2) and has an intercept of 4 units on the positive side of x-axis. The equation of the circle, isA. `x^(2)+y6(2)-4(sqrt(2)x+y)+4=0`B. `x^(2)+y^(2)-4(x+sqrt(2)y)+4=0`C. `x^(2)+y^(2)-2(sqrt(2)x+y)+4=0`D. none of these
If (1, 2) is a limiting point of a coaxial system of circles containing the circle `x^(2)+y^(2)+x-5y+9=0`, then the equation of the radical axis, isA. `x+2y+9=0`B. `3x-y+4=0`C. `x+9y-4=0`D. `3x-y-1=0`
The equation of the circle passing through (0, 0) and belonging to the system of circles of which (3, 1) and (-1, 5) are limiting points, isA. `x^(2)+y^(2)-x+3y=0`B. `x^(2)+y^(2)-11x+3y=0`C. `x^(2)+y^(2)=1`D. none of these
One of the limit point of the coaxial system of circles containing `x^(2)+y^(2)-6x-6y+4=0, x^(2)+y^(2)-2x-4y+3=0`, isA. (-1, 1)B. (-1, 2)C. (-2, 1)D. (-2, 2)
The limiting points of the system of circles represented by the equation `2(x^(2)+y^(2))+lambda x+(9)/(2)=0`, areA. `(pm(3)/(2), 0)`B. `(0, 0), ((9)/(2), 0)`C. `(pm (9)/(2), 0)`D. `(pm 3, 0)`
A ray of light incident at the point (-2, -1) gets reflected from the tangent at (0, -1) to the circle `x^2 +y^2=1`. The reflected ray touches the circle. The equation of the line along which the incident ray moved isA. `4x-3y+11=0`B. `4x+3y+11=0`C. `3x+4y+11=0`D. none of these
Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre.Given: In a circle with centre O, PO is radius and PQ is its chord, seg OR ⊥ chord PQ, P-R-Q, PQ = 24 cm, diameter (d) = 26 cm To Find: Distance of the chord from the centre (OR)