MHT CET · Maths · Circle
If \(y=2 x\) is a chord of circle \(x^2+y^2-10 x=0\), then the equation of circle with this chord as diameter is
- A \(x^2+y^2-2 x-4 y=0\)
- B \(x^2+y^2+2 x+4 y=0\)
- C \(x^2+y^2-2 x+4 y=0\)
- D \(x^2+y^2+2 x-4 y=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-2 x-4 y=0\)
Step-by-step Solution
Detailed explanation
\(x^2-10 x+y^2=0 \)
\( x-10 x+25+y^2=25 \Rightarrow \text { centre }=(5,0) \text { and }\)\(r=5\)
\(\mathrm{y}=2 \mathrm{x}\) is a chord of given circle.
Point of intersection of chord and circle is
\(\mathrm{x}^2-10 \mathrm{x}+25+4 \mathrm{x}^2=25 \quad \Rightarrow \mathrm{y}=0,4\)
Thus end points of the chord are \((0,0)\) and \((2,4)\)
Mid point of the chord \(=\left(\frac{2}{2}, \frac{4}{2}\right)=(1,2)\) and
length of chord \(=\sqrt{(2)^2+(4)^2}=\sqrt{20}\) is the diameter of required circle.
Hence equation of required circle is \((x-1)+(y-2)=\left(\frac{\sqrt{20}}{2}\right)^2\) i.e. \(x^2+y^2-2 x-4 y=0\)
\( x-10 x+25+y^2=25 \Rightarrow \text { centre }=(5,0) \text { and }\)\(r=5\)
\(\mathrm{y}=2 \mathrm{x}\) is a chord of given circle.
Point of intersection of chord and circle is
\(\mathrm{x}^2-10 \mathrm{x}+25+4 \mathrm{x}^2=25 \quad \Rightarrow \mathrm{y}=0,4\)
Thus end points of the chord are \((0,0)\) and \((2,4)\)
Mid point of the chord \(=\left(\frac{2}{2}, \frac{4}{2}\right)=(1,2)\) and
length of chord \(=\sqrt{(2)^2+(4)^2}=\sqrt{20}\) is the diameter of required circle.
Hence equation of required circle is \((x-1)+(y-2)=\left(\frac{\sqrt{20}}{2}\right)^2\) i.e. \(x^2+y^2-2 x-4 y=0\)
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