**Question 1. What does the equation (x â€“ a)**^{2} + (y â€“ b)^{2} = r^{2} become when the axes are transferred to parallel axes through the point (aâ€“c, b)?

^{2}+ (y â€“ b)

^{2}= r

^{2}become when the axes are transferred to parallel axes through the point (aâ€“c, b)?

**Solution:**

We are given,

(x â€“ a)

^{2}+ (y â€“ b)^{2}= r^{2}Putting x = X + a â€“ c and y = Y + b, we get,

=> ((X + a â€“ c) â€“ a)

^{2}+ ((Y + b ) â€“ b)^{2}= r^{2}=> (X â€“ c)

^{2}+ Y^{2}= r^{2}=> X

^{2}+ c^{2}â€“ 2cX + Y^{2}= r^{2}=> X

^{2}+ Y^{2}â€“ 2cX = r^{2}â€“ c^{2}

Therefore the required equation is X^{2}+ Y^{2}â€“2cX = r^{2}â€“ c^{2}.

**Question 2. What does the equation (a â€“ b) (x**^{2} + y^{2}) â€“ 2abx = 0 become if the origin is shifted to the point (ab/(aâ€“b), 0) without rotation?

^{2}+ y

^{2}) â€“ 2abx = 0 become if the origin is shifted to the point (ab/(aâ€“b), 0) without rotation?

**Solution:**

We are given,

(a â€“ b) (x

^{2}+ y^{2}) â€“ 2abx = 0Putting x = X + [ab/(aâ€“b)] and y = Y, we get,

=>

=>

=>

=>

=> X

^{2 }(aâ€“b)^{2 }+ (ab)^{2 }+ 2abX (aâ€“b)+Y^{2 }(aâ€“b)^{2 }= 2abX (aâ€“b)+2(ab)^{2}=> (a â€“ b)

^{2}(X^{2}+ Y^{2}) = a^{2}b^{2}

Therefore the required equation is (a â€“ b)^{2}(X^{2}+ Y^{2}) = a^{2}b^{2}.

**Question 3. Find what the following equations become when the origin is shifted to the point (1, 1)?**

**(i) x**^{2} + xy â€“ 3x â€“ y + 2 = 0

^{2}+ xy â€“ 3x â€“ y + 2 = 0

**Solution:**

We are given,

x

^{2}+ xy â€“ 3x â€“ y + 2 = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)

^{2}+ (X + 1) (Y + 1) â€“ 3(X + 1) â€“ (Y + 1) + 2 = 0=> X

^{2}+ 1 + 2X + XY + X + Y + 1 â€“ 3X â€“ 3 â€“ Y â€“ 1 + 2 = 0

Therefore the required equation is X^{2}+ XY = 0.

**(ii) x**^{2} â€“ y^{2} â€“ 2x + 2y = 0

^{2}â€“ y

^{2}â€“ 2x + 2y = 0

**Solution:**

We are given,

x

^{2}â€“ y^{2}â€“ 2x + 2y = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)

^{2}â€“ (Y + 1)^{2}â€“ 2(X + 1) + 2(Y + 1) = 0=> X

^{2}+ 1 + 2X â€“ Y^{2}â€“ 1 â€“ 2Y â€“ 2X â€“ 2 + 2Y + 2 = 0=> X

^{2}â€“ Y^{2}= 0

Therefore the required equation is X^{2}â€“ Y^{2}= 0.

**(iii) xy â€“ x â€“ y + 1 = 0**

**Solution:**

We are given,

xy â€“ x â€“ y + 1 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) â€“ (X + 1) â€“ (Y + 1) + 1 = 0

=> XY + X + Y + 1 â€“ X â€“ 1 â€“ Y â€“ 1 + 1 = 0

=> XY = 0

Therefore the required equation is XY = 0.

**(iv) xy â€“ y**^{2} â€“ x + y = 0

^{2}â€“ x + y = 0

**Solution:**

We are given,

xy â€“ y

^{2}â€“ x + y = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) â€“ (Y + 1)

^{2}â€“ (X + 1) + (Y + 1) = 0=> XY + X + Y + 1 â€“ Y

^{2}â€“ 1 â€“ 2Y â€“ X â€“ 1 + Y + 1 = 0=> XY â€“ Y

^{2}= 0

Therefore, the required equation is XY â€“ Y^{2}= 0.

**Question 4. At what point the origin be shifted so that the equation x**^{2} + xy â€“ 3x â€“ y + 2 = 0 does not contain any first degree term and constant term?

^{2}+ xy â€“ 3x â€“ y + 2 = 0 does not contain any first degree term and constant term?

**Solution:**

We are given,

x

^{2}+ xy â€“ 3x â€“ y + 2 = 0Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)

^{2}+ (X + a)(Y + b) â€“ 3(X + a) â€“ (Y + b) + 2 = 0=> X

^{2}+ a^{2}+ 2aX + XY + aY + bX + ab â€“ 3X â€“ 3a â€“ Y â€“ b + 2 = 0=> X

^{2}+ XY + X(2a + b â€“ 3) + Y(a â€“ 1) + a^{2}+ ab â€“ 3a â€“ b + 2 = 0As our transformed equation has no first-degree term, we have,

2a + b â€“ 3 = 0 and a â€“ 1 = 0

By solving these equations we have a = 1 and b = 1.

Therefore, the origin has been shifted to (1,1) from (0,0).

**Question 5. Verify that the area of the triangle with vertices (2, 3), (5, 7)**,** and (â€“3, â€“1) remains invariant under the translation of axes when the origin is shifted to the point (â€“1, 3).**

**Solution:**

Here, L.H.S. = A

_{1 }= Area of the triangle with vertices (2, 3), (5, 7) and (â€“3, â€“1)=

=

=

=

= 4 sq. units

As the origin shifted to point (â€“1, 3), the new coordinates of the triangle are:

(X

_{1}, Y_{1}) = (2â€“1, 3+3) = (1, 6)(X

_{2}, Y_{2}) = (5â€“1, 7+3) = (4, 10)(X

_{3}, Y_{3}) = (â€“3â€“1, â€“1+3) = (â€“4, 2)Now, R.H.S. = A

_{2}= Area of the triangle with vertices (1, 6), (4, 10) and (â€“4, 2)=

=

=

=

= 4 sq. units

Therefore, A

_{1}= A_{2}.

Hence, proved.

**Question 6. Find what the following equations become when the origin is shifted to the point (1, 1)?**

**(i) x**^{2} + xy â€“ 3y^{2} â€“ y + 2 = 0

^{2}+ xy â€“ 3y

^{2}â€“ y + 2 = 0

**Solution:**

We are given,

x

^{2}+ xy â€“ 3y^{2}â€“ y + 2 = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)

^{2}+ (X + 1) (Y + 1) â€“ 3(Y + 1)^{2}â€“ (Y + 1) + 2 = 0=> X

^{2}+ 1 + 2X + XY + X + Y + 1 â€“ 3Y^{2}â€“ 3 â€“ 6Y â€“ Y â€“ 1 + 2 = 0=> X

^{2}â€“ 3Y^{2}+ XY + 3X â€“ 6Y = 0

Therefore, the required equation is X^{2}â€“ 3Y^{2}+ XY + 3X â€“ 6Y = 0.

**(ii)** **xy â€“ y**^{2} â€“ x + y = 0

^{2}â€“ x + y = 0

**Solution:**

We are given,

xy â€“ y

^{2}â€“ x + y = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) â€“ (Y + 1)

^{2}â€“ (X + 1)+ Y + 1 = 0=> XY + X + Y + 1 â€“ Y

^{2}â€“ 1 â€“ 2Y â€“ X â€“ 1 + Y + 1 = 0=> XY â€“ Y

^{2}= 0

Therefore, the required equation is XY â€“ Y^{2}= 0.

**(iii) xy â€“ x â€“ y + 1 = 0**

**Solution:**

We are given,

xy â€“ x â€“ y + 1 = 0

Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1) (Y + 1) â€“ (Y + 1) â€“ (X + 1) + 1 = 0

=> XY + X + Y + 1 â€“ Y â€“ 1 â€“ X â€“ 1 + 1 = 0

=> XY = 0

Therefore, the required equation is XY = 0.

**(iv) x**^{2} â€“ y^{2} â€“ 2x + 2y = 0

^{2}â€“ y

^{2}â€“ 2x + 2y = 0

**Solution:**

We are given,

x

^{2}â€“ y^{2}â€“ 2x + 2y = 0Putting x = X + 1 and y = Y + 1, we get,

=> (X + 1)

^{2}â€“ (Y + 1)^{2}â€“ 2(X + 1) + 2(Y + 1) = 0=> X

^{2}+ 1 + 2X â€“ Y^{2}â€“ 1 â€“ 2Y â€“ 2X â€“ 2 + 2Y + 2 = 0=> X

^{2}â€“ Y^{2}= 0

Therefore, the required equation is X^{2}â€“ Y^{2}= 0.

**Question 7. Find the point to which the origin should be shifted after the translation of axes so that the following equations will have no first degree terms.**

**(i) x**^{2} + y^{2} â€“ 4x â€“ 8y + 3 = 0

^{2}+ y

^{2}â€“ 4x â€“ 8y + 3 = 0

**Solution:**

We are given,

x

^{2}+ y^{2}â€“ 4x â€“ 8y + 3 = 0Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)

^{2}+ (Y + b)^{2}â€“ 4(X + a) â€“ 8(Y + b) + 3 = 0=> X

^{2}+ a^{2}+ 2aX + Y^{2}+ b^{2}+ 2bY â€“ 4X â€“ 4a â€“ 8Y â€“ 8b + 3 = 0=> X

^{2}+ Y^{2}+ (2a â€“ 4)X + (2b â€“ 8)Y + (a^{2}+ b^{2}â€“ 4a â€“ 8b +3) = 0As our transformed equation has no first-degree term, we have,

2a â€“ 4 = 0 and 2b â€“ 8 = 0

By solving these equations we have a = 2 and b = 4.

Therefore, the origin has been shifted to (2,4) from (0,0).

**(ii) x**^{2} + y^{2} â€“ 5x + 2y â€“ 5 = 0

^{2}+ y

^{2}â€“ 5x + 2y â€“ 5 = 0

**Solution:**

We are given,

x

^{2}+ y^{2}â€“ 5x + 2y â€“ 5 = 0Suppose (a, b) is the point where the origin has been shifted from (0, 0). Putting x = X + a and y = Y + b, we get the transformed equation,

=> (X + a)

^{2}+ (Y + b)^{2}â€“ 5(X + a) + 2(Y + b) â€“ 5 = 0=> X

^{2}+ a^{2}+ 2aX + Y^{2}+ b^{2}+ 2bY â€“ 5X â€“ 5a + 2Y + 2b â€“ 5 = 0=> X

^{2}+ Y^{2}+ (2a â€“ 5)X + (2b + 2)Y + (a^{2}+ b^{2}â€“ 5a + 2b â€“ 5) = 0As our transformed equation has no first-degree term, we have,

2a â€“ 5 = 0 and 2b + 2 = 0

By solving these equations we have a = 5/2 and b = â€“1.

Therefore, the origin has been shifted to (5/2, â€“1) from (0,0).

**(iii) x**^{2} â€“ 12x + 4 = 0

^{2}â€“ 12x + 4 = 0

**Solution:**

We are given,

x

^{2}â€“ 12x + 4 = 0=> (X + a)

^{2}â€“ 12(X + a) + 4 = 0=> X

^{2}+ a^{2}+ 2aX â€“ 12X â€“ 12a + 4 = 0=> X

^{2}+ (2a â€“ 12)X + (a^{2 }â€“ 12a + 4) = 0As our transformed equation has no first-degree term, we have,

=> 2a â€“ 12 = 0

=> a = 6

Therefore, the origin has been shifted to (6,b) from (0,0) where b is any arbitrary value.

**Question 8. Verify that the area of the triangle with vertices (4, 6), (7, 10)**,** and (1, â€“2) remains invariant under the translation of axes when the origin is shifted to the point (â€“2, 1).**

**Solution:**

Here, L.H.S. = A

_{1}= Area of the triangle with vertices (4, 6), (7, 10) and (1, â€“2)=

=

=

=

= 6 sq. units

As the origin shifted to point (â€“2, 1), the new coordinates of the triangle are:

(X

_{1}, Y_{1}) = (4â€“2, 6+1) = (2, 7)(X

_{2}, Y_{2}) = (7â€“2, 10+1) = (5, 11)(X

_{3}, Y_{3}) = (1â€“2, â€“2+1) = (â€“1, â€“1)Now, R.H.S. = A

_{2}= Area of the triangle with vertices (2, 7), (5, 11), (â€“1, â€“1)=

=

=

=

= 6 sq. units

Therefore, A

_{1}= A_{2}.

Hence, proved.