Question

If a chord of circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$=8 makes equal intercepts of length ‘a’ on the coordinate axes then |a|<...

Answer 1

Hint: To solve this question, here we use basic theories of chapter circles. As mentioned in this question, the chord makes equal intercepts of length a on the coordinate axes so by using this statement we write the general equation of chord to the circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$= 8. And as we know, the length of the perpendicular from the center of the given circle to the chord must be less than the radius.

Complete step-by-step answer:
Given in question,
Circle having equation:
${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$= 8
That means circle having centre (0, 0) and radius = $\sqrt {\text{8}}$=$2\sqrt 2$
Since the chord makes equal intercepts of length a on the coordinate aces.
So, its equation can be written as x±y =±a
This line meets the given circle at two distinct points.
So, the length of the perpendicular from the center (0,0) of the given circle must be less than the radius.

And in this case,
$\left| {{\text{ }}\dfrac{{\text{a}}}{{\sqrt {\text{2}} }}} \right|{< \sqrt 8 }$
Taking square on both the side,
⇒ ${{\text{a}}^{\text{2}}}$<16
⇒ $\left| {\text{a}} \right|$<4
Therefore, If a chord of circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}$=8 makes equal intercepts of length ‘a’ on the coordinate axes then |a|< 4.

Note- Chords are equidistant from the center if and only if their lengths are equal. Equal chords are subtended by equal angles from the center of the circle. A chord that passes through the center of a circle is called a diameter and is the longest chord.