Question

Determine equation of the circle whose diameter is the chord x+y=1 of the circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ = 4}}$

Answer 1

Hint: To solve this question, we use basic theory of circles. In this first we find the two points on the circle by using given chord equation and then after using these two points we easily Determine equation of the circle whose diameter is the chord x+y=1 of the circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ = 4}}$.

Complete step-by-step answer:
As given in question,
Circle: ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ = 4}}$ …………….. (1)
Chord: x+y=1
y=1-x ………………(2)

                                             ![](https://www.vedantu.com/question-sets/339acbdb-f167-4b82-9d67-ec2603a2ff335123491898128391922.png)  

now, put value of y in equation (1), we get
${{\text{x}}^{\text{2}}}{\text{ + (1 - x}}{{\text{)}}^{\text{2}}}{\text{ = 4}}$
${{\text{x}}^{\text{2}}}{\text{ + 1 + }}{{\text{x}}^{\text{2}}}{\text{ - 2x = 4}}$
2${{\text{x}}^{\text{2}}}$-2x-3=0
x= $\dfrac{{{ - b \pm }\sqrt {\text{D}} }}{{{\text{2a}}}}$
by using this we get,
x= $\dfrac{{{1 \pm }\sqrt {\text{7}} }}{{\text{2}}}$ (${{\text{x}}_{\text{1}}}{\text{ and }}{{\text{x}}_{\text{2}}}$)
and by putting value of x, we get value of y
y= ${\text{1 - }}\left( {\dfrac{{{1 \pm }\sqrt {\text{7}} }}{{\text{2}}}} \right)$
y= $\dfrac{1}{2} - \dfrac{{\sqrt 7 }}{2}$or $\dfrac{1}{2} + \dfrac{{\sqrt 7 }}{2}$ (${{\text{y}}_{\text{1}}}{\text{ and }}{{\text{y}}_{\text{2}}}$)
So, when x= $\dfrac{{{\text{1 + }}\sqrt {\text{7}} }}{{\text{2}}}$ $\Rightarrow$ y= $\dfrac{1}{2} - \dfrac{{\sqrt 7 }}{2}$
And x= $\dfrac{{{\text{1 - }}\sqrt {\text{7}} }}{{\text{2}}}$ $\Rightarrow$ y= $\dfrac{1}{2} + \dfrac{{\sqrt 7 }}{2}$
Now, equation of new circle passing through points A and B.
$\left( {{\text{x - }}{{\text{x}}_{\text{1}}}} \right){\text{(x - }}{{\text{x}}_{\text{2}}}{\text{) + (y - }}{{\text{y}}_{\text{1}}}{\text{)(y - }}{{\text{y}}_{\text{2}}}{\text{) = }}$0
${{\text{x}}^{\text{2}}}{\text{ - x(}}{{\text{x}}_{\text{1}}}{\text{ + }}{{\text{x}}_{\text{2}}}{\text{) + }}{{\text{x}}_{\text{1}}}{{\text{x}}_{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - y(}}{{\text{y}}_{\text{1}}}{\text{ + }}{{\text{y}}_{\text{2}}}{\text{) + }}{{\text{y}}_{\text{1}}}{{\text{y}}_{\text{2}}}{\text{ = 0}}$
Here put all values of x and y, which we calculated above.
${{\text{x}}_{\text{1}}}{\text{ + }}{{\text{x}}_{\text{2}}}$= 1
${{\text{x}}_{\text{1}}}{{\text{x}}_{\text{2}}}$= $\dfrac{{ - 3}}{2}$
${{\text{y}}_{\text{1}}}{\text{ + }}{{\text{y}}_{\text{2}}}$= 1
${{\text{y}}_{\text{1}}}{{\text{y}}_{\text{2}}}$= $\dfrac{{ - 3}}{2}$
Now, we get:
${{\text{x}}^{\text{2}}}{\text{ - x(1) + }}\left( {\dfrac{{ - 3}}{2}} \right){\text{ + }}{{\text{y}}^{\text{2}}}{\text{ - y(1) + }}\left( {\dfrac{{ - 3}}{2}} \right){\text{ = 0}}$
${\text{2}}{{\text{x}}^{\text{2}}}{\text{ - 2x - 3 + 2}}{{\text{y}}^{\text{2}}}{\text{ - 2y - 3 = 0}}$
Therefore, equation of the circle whose diameter is the chord x+y=1 of the circle ${{\text{x}}^{\text{2}}}{\text{ + }}{{\text{y}}^{\text{2}}}{\text{ = 4}}$ is ${\text{2}}{{\text{x}}^{\text{2}}}{\text{ - 2x - 3 + 2}}{{\text{y}}^{\text{2}}}{\text{ - 2y - 3 = 0}}$.

Note- A chord is a line passing from one point to another on the circumference of a circle. Diameter is a chord that passes through the centre. The perimeter of a circle is known as the circumference. Chords equal in length subtend equal angles at centre. Similarly, the chords are equal if the angles subtended by chords are equal at the centre. The perpendicular drawn on a chord from the centre bisects the chord.