Question

The geometric means of the abscissa of the points of intersection of the circle $x^2+y^2-4x-6y+7=0$ and the line $y=1$ is
(a)$\sqrt{7}$
(b)$\sqrt{2}$
(c)$\sqrt{14}$
(d) 1

Answer 1

Hint: Find the points of intersection of the line and circle, to find the x-coordinate or abscissa. Then find the geometric means using the formula$\sqrt{ab}$.

Complete step-by-step answer:
In the given question, the equation of the circle is,
$x^2+y^2-4x-6y+7=0$
The equation of line is,
y = 1
We can find the point of intersection of line and the circle by putting ‘y = 1’ in equation of the circle,
$x^2+y^2-4x-6y+7=0$
By substituting, we get
$x^2+{(1)}^2-4x-6(1)+7=0$
$\Rightarrow x^2-4x+2=0$
Now we will convert this into square,
$\Rightarrow x^2-4x+4=2$
$\Rightarrow x^2-2x-2x+4=2$
$\Rightarrow {(x-2)}^2=2$
Taking square on both sides, we get
$$\Rightarrow x-2=\pm \sqrt{2}$$
$$\therefore x=2\pm \sqrt{2}$$
Abscissa represents the value of x-coordinate corresponding to that point.
So, the abscissa of points which we found are $(2+\sqrt{2})$ and $(2-\sqrt{2})$.
In the question we are asked to find the geometric mean of the values.
We know the formula that the geometric mean between two values a,b is $\sqrt{ab}$.
So, geometric mean in this case will be,
$\sqrt{(2+\sqrt{2})(2-\sqrt{2})}$
Now we know the formula, $(a+b)(a-b)=a^2-b^2$ , so the above equation can be written as,
$\sqrt{2^2-{\left(\sqrt{2}\right)}^2}=\sqrt{4-2}=\sqrt{2}$
So, the geometric means of the abscissa of the points of intersection of the circle $x^2+y^2-4x-6y+7=0$ and the line $y=1$ is $\sqrt{2}$ .
Hence, the correct answer is option (b).
Note: In this question, students should be careful while working out and finding the values of x.
Common mistakes made by students are while writing the formula of geometric mean. They find the abscissa and get to stop there itself.