Question

The other end of diameter through the point $( - 1,1)$ on the circle ${x^2} + {y^2} - 6x + 4y - 12 = 0$ is :
(A) $\left( { - 7,5} \right)$
(B) $\left( { - 7, - 5} \right)$
(C) $\left( {7, - 5} \right)$
(D) $\left( {7,5} \right)$

Answer 1

Hint:As the question say that one end point of diameter is $( - 1,1)$ and we have to find other endpoint so for this first we find center of circle by using general equation of circle that is ${x^2} + {y^2} + 2gx + 2fy + c = 0$ So the center of the circle is $( - g, - f)$ and it is the midpoint of diameter so we use section formulas here to find our answer.
$(x,y) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$ (it is used only when ratio is $1:1$ that mean mid point ). Here $(x,y)$ is the center of a given circle.

Complete step-by-step answer:
Equation of circle ${x^2} + {y^2} - 6x + 4y - 12 = 0$
And one endpoint of diameter is $( - 1,1)$
So assume second end point of diameter is $\left( {a,b} \right)$
Now we find center of circle by comparing given circle equation to the ${x^2} + {y^2} + 2gx + 2fy + c = 0$
So from this we get $g = - 3$ and $f = 2$
So center is $\left( { - g, - f} \right)$ = $(3, - 2)$
Now to find the other endpoint we use section formula that is
$(x,y) = \left( {\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2}} \right)$
By putting values we get
 $(3, - 2) = \left( {\dfrac{{ - 1 + a}}{2},\dfrac{{1 + b}}{2}} \right)$
Now $3 = \dfrac{{ - 1 + a}}{2}$ and $ - 2 = \dfrac{{1 + b}}{2}$
Now we solve this one by one
$3 = \dfrac{{ - 1 + a}}{2}$
$6 = - 1 + a$
So $a = 7$
Now we solve $ - 2 = \dfrac{{1 + b}}{2}$
From this we get
$ - 4 = 1 + b$
So $b = - 5$
Form the value of a and b we can say that another point of diameter is $(7, - 5)$

So, the correct answer is “Option C”.

Note:If a point \[R\] lies between $P$ and $Q$ and divide $PQ$ in the ratio of $m:n$ so by using section formula we find points of $R$ So $(x,y) = \left( {\dfrac{{m{x_1} + n{y_1}}}{{m + n}},\dfrac{{m{x_2} + n{y_2}}}{{m + n}}} \right)$.