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TN 12th Computer Applications மின்னணு தரவு பரிமாற்றம் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிக பாதுகாப்பு அமைப்புகள் Sample Question Papers Study Material - QB365 Set A

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12th Standard

TN 12th Computer Applications மின்னணு செலுத்தல் முறைகள் Sample Question Papers Study Material - QB365 Set A

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TN 12th Computer Applications மின் - வணிகம் Sample Question Papers Study Material - QB365 Set A

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12th Standard

TN 12th Computer Applications திறந்த மூல கருத்துருக்கள் Sample Question Papers Study Material - QB365 Set A

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12th Standard

TN 12th Computer Applications வலையமைப்பு வடமிடல் Sample Question Papers Study Material - QB365 Set A

12th Standard English Medium Maths Subject Two Dimensional Analytical Geometry-II Book Back 2 Mark Questions with Solution Part - I

Published on: 01/06/2021

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Std: 12th Standard - Maths

chapter: - Book back Questions

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Questions:

Generic 2

4 x 2 = 8

Examine the position of the point (2, 3) with respect to the circle x²+ y²− 6x − 8y + 12 = 0.

Find the general equation of the circle whose diameter is the line segment joining the points (−4, −2) and (1, 1) is x²+y²+5x+3y+6=0

Determine whether x + y − 1 = 0 is the equation of a diameter of the circle x²+ y²− 6x + 4y + c = 0 for all possible values of c .

Find the general equation of a circle with centre (-3, -4) and radius 3 units.

Generic 3

1 x 3 = 3

The line 3x+4y−12 = 0 meets the coordinate axes at A and B. Find the equation of the circle drawn on AB as diameter.

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Answers Key:

Generic 2

4 x 2 = 8

Taking (x₁,y₁) as (2, 3), we get
x₁² + y₁²+ 2gx₁+ 2fy₁+ c = 2²+ 3²− 6 × 2 − 8 × 3 + 12
= 4 + 9 - 12 - 24 + 12
= -11\(<\)0.
Therefore the point (2, 3) lies inside the circle, by theorem.

Equation of the circle with end points of the diameter as (x₁, y₁₎and (x₂, y₂) given in theorem is
(x−x₁)(x−x₂)+(y−y₁)(y−y₂) = 0
(x+4)(x−1)+(y+2)(y−1) = 0
x²+ y²+ 3x + y − 6 = 0 which is the required equation of the circle.

Centre of the circle is (3,-2) which lies on x + y − 1 = 0. So the line x + y − 1 = 0 passes through the centre and therefore the line x + y −1 = 0 is a diameter of the circle for all possible values of c .

Equation of the circle in standard form is (xr − h)²+ (y − k)²= r²
\( \Rightarrow (x-(-3))^{2}+(y-(-4))^{2} =3^{2} \)
\( \Rightarrow (x+3)^{2}+(y+4)^{2} =3^{2} \)
\( \Rightarrow x^{2}+y^{2}+6 x+8 y+16 =0 .\)

Generic 3

1 x 3 = 3

Writing the line 3x+4y = 12, in intercept form yields \(\frac{x}{4}+\frac{y}{3}=1\). Hence the points A and B are (4, 0) and (0, 3) .
Equation of the circle in diameter form is
(x-x₁) (x-x₂)+(y-y₁) (y-y₂) = 0
(x-4) (x-0)+(y-0) (y-3) = 0
x²+y²−4x−3y = 0 .