The intercept on the line $y = x$ by the circle $x^2 + y^2 - 2x = 0$ is $AB$. Equation of the circle with $AB$ as diameter is
Show Hint
The equation of any circle that passes through the point of intersection
x2 + y2 - 2x = 0 and y = x is
x2 + y2 - 2x + λ(y - x) = 0
⇒ x2 + y2 - (2 + λ)x + λy = 0
Center of the circle is \((\frac{2+λ}{2},\frac{-λ}{2})\)
For AB to be the diameter of the desired circle, its center must be located on line segment AB i.e
\(\frac{2+λ}{2}=\frac{-λ}{2}⇒λ=1\)
Therefore, the equation of required circle is
x2 + y2 - x - y = 0
So, the correct option is (B) : $x(x - 1) +y(y - 1) = 0$
- $x^2 + y^2 = 1$
- $x(x - 1) +y(y - 1) = 0$
- $x^2 + y^2 = 2$
- $(x -1)(x-2)+(y-1)+(y-2)= 0$
The Correct Option is B
Solution and Explanation
The intersection of the line and circle is
\(x^{2}+x^{2}-2 x=0\)
\(\Rightarrow 2 x(x-1)=0\)
\(\Rightarrow x=0,1\)
\(\Rightarrow y=0,1\)
\(\therefore\) The points \((0,0)\) and \((1,1)\) are the end points of a diameter of circle
\(\therefore\) Its equation is \((x-0)(x-1)+(y-0)(y-1)=0\)
\(\Rightarrow x^{2}+y^{2}-x-y=0\)
So, the correct option is (B) : $x(x - 1) +y(y - 1) = 0$
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Concepts Used:
Circle
A circle can be geometrically defined as a combination of all the points which lie at an equal distance from a fixed point called the centre. The concepts of the circle are very important in building a strong foundation in units likes mensuration and coordinate geometry. We use circle formulas in order to calculate the area, diameter, and circumference of a circle. The length between any point on the circle and its centre is its radius.
Any line that passes through the centre of the circle and connects two points of the circle is the diameter of the circle. The radius is half the length of the diameter of the circle. The area of the circle describes the amount of space that is covered by the circle and the circumference is the length of the boundary of the circle.
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