Question

The centres of the circles \[{x^2} + {y^2} = 1,{x^2} + {y^2} = 6x - 2y - 1 = 0\] and \[{x^2} + {y^2} - 12x + 4y - 1 = 0\] are
A. The vertices of an equilateral triangle
B. The vertices of a right angled triangle
C. The vertices of an isosceles triangle
D. Points are collinear

Answer 1

Hint: Here we have a center point of the circle and the circle equation \[{x^2} + {y^2} = 1,{x^2} + {y^2} = 6x - 2y - 1 = 0\] and \[{x^2} + {y^2} - 12x + 4y - 1 = 0\]. We have to calculate the distance between each circle using the formula\[\sqrt {{{({x_2} - {x_1})}^2} - {{({y_2} - {y_1})}^2}} \] and then find the solution for each circle and find the collinearity.

Complete step by step Solution:
equation. In the given equation, we will obtain the coordinate axis.
The center of the circle \[{x^2} + {y^2} = 1\]is \[O(0,0)\]
The circle is of the form \[\left. {{{(x - a)}^2} + {{(y - b)}^2} = {r^2}} \right]\]
The center of the circle\[{x^2} + {y^2} + 6x - 2y = 1\] is \[A( - 3,1)\]
Because, the circle is of the form\[{x^2} + {y^2} + 2fx + 2gy + c = 0\]
Then the center is\[(f, - g)\].
The center of the circle\[{x^2} + {y^2} - 12x + 4y = 1\] is \[B(6, - 2)\]
Because, the circle is of the form\[{x^2} + {y^2} + 2fx + 2gy + c = 0\]
We have to first determine the distance between the first two circles:
\[\therefore OA = \sqrt {{{( - 3 - 0)}^2} + {{(1 - 0)}^2}} \]
Simplify the terms inside the root:
\[ = \sqrt {9 + 1} \]
We have to add the terms inside the root:
\[ = \sqrt {10} \]
Now we have to determine the distance between the next two circles\[{\rm{AB}}\]:
Substitute the values in the formula:
\[AB = \sqrt {{{(6 + 3)}^2} + {{( - 2 - 1)}^2}} \]
Simplify the terms inside the root:
\[ = \sqrt {81 + 9} \]
Add the terms inside the bracket:
\[ = \sqrt {90} \]
Prime factorization of\[90\]:
\[ = \sqrt {2 \cdot {3^2} \cdot 5} \]:
Apply radical rule\[\sqrt {ab} = \sqrt a \sqrt b \]:
\[ = \sqrt {{3^2}} \sqrt {2 \cdot 5} \]
Apply radical rule\[\sqrt {{a^2}} = a\]:
\[ = 3\sqrt {2 \cdot 5} \]
Multiply the numbers\[2 \cdot \:5 = 10\]:
\[ = 3\sqrt {10} \]
Now we have to determine the distance between the next two circles\[{\rm{OB}}\]:
\[OB = \sqrt {{{(6 - 0)}^2} + {{( - 2 - 0)}^2}} \]
Solve the terms inside the root:
\[ = \sqrt {36 + 4} \]
Add the terms:
\[ = \sqrt {40} \]
Prime factorization of\[40\]:
\[ = \sqrt {{2^3} \cdot \:5} \]
Apply exponent rule\[{2^3} \cdot 5 = {2^2} \cdot 2 \cdot 5:\]
\[ = \sqrt {{2^2} \cdot \:2 \cdot \:5} \]
Apply radical rule \[\sqrt {{2^2} \cdot \:2 \cdot \:5} = \sqrt {{2^2}} \sqrt {2 \cdot \:5} \]:
\[ = \sqrt {{2^2}} \sqrt {2 \cdot \:5} \]
Apply radical rule\[\sqrt {{2^2}} = 2\]:
\[ = 2\sqrt {2 \cdot \:5} \]
Multiply the numbers\[2 \cdot \:5 = 10\]
:\[ = 2\sqrt {10} \]
Now,\[AO + OB = \sqrt {10} + 2\sqrt {10} \]
Add similar elements:
\[ = 3\sqrt {10} = AB\]
Therefore, \[A,{\rm{ }}O,{\rm{ }}B\] lie on a straight line.
The points are collinear.
Therefore, the correct option is (D).

Note: There are two circles in this question. This question has a single center point for a single circle. First, we'll find another circle's centre point. This time, we'll have a firm grasp on the center points. Then we have to find the distance between each circle to find the point that lies in a straight line.