Question

A rod AB of length $4$ units moves horizontally when its left end A always on the circle ${x^2} + {y^2} - 4x - 18y - 29 = 0$, then the locus of the other end B is
A ${x^2} + {y^2} - 12x - 8y + 3 = 0$
B ${x^2} + {y^2} - 12x - 8y + 3 = 0$
C ${x^2} + {y^2} + 4x - 8y - 29 = 0$
D ${x^2} + {y^2} - 4x - 16y + 19 = 0$

Answer 1

Hint: Given, a rod AB of length $4$units moves horizontally when its left end A always on the circle ${x^2} + {y^2} - 4x - 18y - 29 = 0$, First we will find the center of the circle and the radius of the circle. Then we will find the center and radius of the locus of B. With the help of the new center and radius to get the equation locus of B.

Complete step by step solution:
${(x - {x_0})^2} + {(y - {y_0})^2} = {r^2}$
Complete step by step solution: Given, a rod AB of length $4$units moves horizontally when its left end A always on the circle ${x^2} + {y^2} - 4x - 18y - 29 = 0$
Equation of circle ${x^2} + {y^2} - 4x - 18y - 29 = 0$
Adding $4,81$ on both sides
${x^2} + {y^2} - 4x - 18y - 29 + 4 + 81 = 81 + 4$
Further solving above equation
${(x - 2)^2} + {(x - 3)^2} - 29 = 85$
Shifting constant term on one side
${(x - 2)^2} + {(x - 3)^2} = 114$
Equation of circle
${(x - {x_0})^2} + {(y - {y_0})^2} = {r^2}$
Comparing with this equation
${x_0} = 2,\,{y_0} = 9,\,r = \sqrt {114} $
So, Center of the circle $(2,9)$ and radius is $\sqrt {114} $
A rod of 4 units length with its left end and A is always on the given circle.
Then, the radius of the new circle will be the same. As it moves horizontally, the center of the new circle shifts by 4 units in x coordinates.
Hence, center of new circle is $(2 + 4,9)$
On simplifying center is $(6,9)$
Radius will be same $\sqrt {114} $
Now, the locus of B is the equation of new circle having center $(6,9)$ and radius $\sqrt {114} $
General equation of circle
${(x - {x_0})^2} + {(y - {y_0})^2} = {r^2}$
Now, ${x_0} = 6,\,{y_0} = 9,\,r = \sqrt {114} $
Putting these values in above equation
${(x - 6)^2} + {(y - 9)^2} = {\left( {\sqrt {114} } \right)^2}$
Using this identity ${(a - b)^2} = {a^2} + {b^2} - 2ab$
${x^2} + 36 - 12x + {y^2} + 81 - 18y = 114$
After simplifying above equation
${x^2} - 12x + {y^2} - 18y + 117 = 114$
Shifting variables on one sides and constants on other side
${x^2} - 12x + {y^2} - 18y = 114 - 117$
After simplifying the above equation
${x^2} - 12x + {y^2} - 18y = - 3$
Shifting $ - 3$ on the other side
${x^2} + {y^2} - 12x - 18y + 3 = 0$

Option ‘A’ is correct

Note: Remember that locus is nothing but another circle of same radius as the given circle