
One of the diameters of the circle ${x^2} + {y^2} - 12x + 4y + 6 = 0$is given by
$
(a){\text{ x + y = 0}} \\
(b){\text{ x + 3y = 0}} \\
(c){\text{ x = y}} \\
(d){\text{ 3x + 2y = 0}} \\
$
Answer
593.1k+ views
Hint: In this question we have been given an equation of circle and we need to find equation of its diameter, firstly convert the given equation into standard of equation ${\left( {x - g} \right)^2} + {\left( {y - f} \right)^2} = {r^2}$where g, f are the center points and r is the radius of circle. Then use the property of diameter that it always passes through the center, so directly use the options given to see which of them satisfies.
Complete step-by-step answer:
${x^2} + {y^2} - 12x + 4y + 6 = 0$
Convert this equation into standard form of the circle which is,
${\left( {x - g} \right)^2} + {\left( {y - f} \right)^2} = {r^2}$
Where (g, f) and (r) is the center and radius of the circle respectively.
So make the complete square in x and y respectively by adding and subtracting by square of half of coefficients of x and y respectively.
Therefore the coefficient of x is (-12) so half of it is (-6) so square of it (36).
And the coefficient of y is (4) so half of it is (2) so square of it (4).
Therefore the equation becomes
$ \Rightarrow {x^2} + {y^2} - 12x + 4y + 6 + 36 - 36 + 4 - 4 = 0$
Now simplify the above equation we have,
$ \Rightarrow {\left( {x - 6} \right)^2} + {\left( {y + 2} \right)^2} = 34$
So on comparing the value of g is (6) and the value of f is (-2)
Therefore the center of the circle is (6, -2).
Now as we know that the diameter of the circle is always passing through the center of the circle so the center of the circle is always satisfying the equation of the circle.
So check all the options which satisfy the center of the circle.
So check option (A)
$
\Rightarrow x + y = 0 \\
\Rightarrow 6 - 2 = 4 \ne 0 \\
$
So option (A) is incorrect.
Now check option (B)
$
\Rightarrow x + 3y = 0 \\
\Rightarrow 6 + 3\left( { - 2} \right) = 6 - 6 = 0 \\
$
So option (B) is correct.
And rest two options are also incorrect.
Therefore the equation of the diameter of the circle is $\left( {x + 3y = 0} \right)$.
Hence option (B) is correct.
Note: Whenever we face such types of problems the key concept is to have a good understanding of the basic general equations of figures like circle, moreover directly checking the options form the questions always helps in saving a lot of time in competitive examinations, so work smart . These concepts will help you get on the right track to find the correct answer.
Complete step-by-step answer:
${x^2} + {y^2} - 12x + 4y + 6 = 0$
Convert this equation into standard form of the circle which is,
${\left( {x - g} \right)^2} + {\left( {y - f} \right)^2} = {r^2}$
Where (g, f) and (r) is the center and radius of the circle respectively.
So make the complete square in x and y respectively by adding and subtracting by square of half of coefficients of x and y respectively.
Therefore the coefficient of x is (-12) so half of it is (-6) so square of it (36).
And the coefficient of y is (4) so half of it is (2) so square of it (4).
Therefore the equation becomes
$ \Rightarrow {x^2} + {y^2} - 12x + 4y + 6 + 36 - 36 + 4 - 4 = 0$
Now simplify the above equation we have,
$ \Rightarrow {\left( {x - 6} \right)^2} + {\left( {y + 2} \right)^2} = 34$
So on comparing the value of g is (6) and the value of f is (-2)
Therefore the center of the circle is (6, -2).
Now as we know that the diameter of the circle is always passing through the center of the circle so the center of the circle is always satisfying the equation of the circle.
So check all the options which satisfy the center of the circle.
So check option (A)
$
\Rightarrow x + y = 0 \\
\Rightarrow 6 - 2 = 4 \ne 0 \\
$
So option (A) is incorrect.
Now check option (B)
$
\Rightarrow x + 3y = 0 \\
\Rightarrow 6 + 3\left( { - 2} \right) = 6 - 6 = 0 \\
$
So option (B) is correct.
And rest two options are also incorrect.
Therefore the equation of the diameter of the circle is $\left( {x + 3y = 0} \right)$.
Hence option (B) is correct.
Note: Whenever we face such types of problems the key concept is to have a good understanding of the basic general equations of figures like circle, moreover directly checking the options form the questions always helps in saving a lot of time in competitive examinations, so work smart . These concepts will help you get on the right track to find the correct answer.
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