Question

If $\left( {6, - 3} \right)$ is the one extremity of diameter to the circle ${x^2} + {y^2} - 3x + 8y - 3 = 0$ then its other extremity is

Answer 1

Hint: In this Question, we have to basically use the general equation of the circle to compare it with the given equation to find coordinates of the centre and then look for the other extremity which is pretty simple by the formula of midpoint. By using these basics, we should be able to solve the question.

Complete step-by-step answer:

The given circle is
${x^2} + {y^2} - 3x + 8y - 3 = 0$
The general equation of a circle is
${x^2} + {y^2} + 2gx + 2fy + c = 0$
Comparing the given equation with this equation
$2g = - 3 \Rightarrow g = \dfrac{{ - 3}}{2} \\\ 2f = 8 \Rightarrow f = 4 \\\$
Therefore, the centre of the equation is given by
$\left( { - g, - f} \right) \\\ = \left( {\dfrac{3}{2}, - 4} \right) \\\$

Now, we know that centre is the midpoint of the two extremities of diameter.
Let the other extremity of diameter is $\left( {x,y} \right)$.
Thus, centre is the midpoint $\left( {x,y} \right)$ and $\left( {6, - 3} \right)$

Formula of midpoint is given by
$\left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
Therefore, we have
$\dfrac{{x + 6}}{2} = \dfrac{3}{2} \\\ x + 6 = 3 \\\ x = - 3 \\\$
And
$\dfrac{{y - 3}}{2} = - 4 \\\ y - 3 = - 8 \\\ y = - 5 \\\$

Therefore, the other extremity of diameter is $\left( { - 3, - 5} \right)$

Hence, (B) is the correct option.

Note: While comparing a general formula with a given equation, take precautions with negative signs. Because, the change in sign will lead us to the wrong approach towards solutions.