Single Choice

If a circle passing through the point $$(-1, 0)$$ touches y-axis at $$(0, 2)$$, then the length of the chord of the circle along the x-axis is

A$$\displaystyle \frac{5}{2}$$
B$$5$$
C$$\displaystyle \frac{3}{2}$$
D$$3$$
Correct Answer

Solution

Since the required circle touches y-axis at $$(0,2)$$ and let $$r$$ be the radius of the circle.
So, the center will be $$(-r,2)$$.

Equation of circle is
$$(x+r)^2+(y-2)^2=r^2$$

Since, it passes through $$(-1,0)$$,

$$\Rightarrow (r-1)^2+4=r^2$$

$$\Rightarrow r=\dfrac{5}{2}$$

So the center is $$(-\dfrac{5}{2},2)$$

Now, the perpendicular from the center of the circle to the chord bisects the chord.

So, $$AB=2AM$$

Coordinates of $$M$$ is $$(-\dfrac{5}{2},0)$$

So, $$AM=\dfrac{3}{2}$$

$$\Rightarrow AB=3$$

SIMILAR QUESTIONS

Circles

Show that the equation of a straight line meeting the circle $$x^2+y^2=a^2$$ in two points at equal distance 'd' from a point $$(x_1, y_1)$$ on its circumference is $$xx_1+yy_1-a^2+\left(\dfrac{d^2}{2}\right)=0$$.

Circles

The locus of midpoints of the chords of contact of $$x^2 + y^2 = 2$$ from the points on the line $$3x + 4y = 10$$ is a circle with centre $$P$$. If $$O$$ be the origin, then $$OP$$ is equal to

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