MHT CET · Maths · Circle
The equation of a circle which has a tangent \(3 x+4 y=6\) and two normals given by \((x-1)(y-2)=0\) is
- A \((x-3)^{2}+(y-4)^{2}=5^{2}\)
- B \(x^{2}+y^{2}-4 x-2 y+4=0\)
- C \(x^{2}+y^{2}-2 x-4 y+4=0\)
- D \(x^{2}+y^{2}-2 x-4 y+5=0\)
Answer & Solution
Correct Answer
(C) \(x^{2}+y^{2}-2 x-4 y+4=0\)
Step-by-step Solution
Detailed explanation
\((x-1)(y-2)=0\)
\(\Rightarrow \quad x-1=0 \text { and } y-2=0\)
\(\therefore \text {Radius}=\frac{3(1)+4(2)-6}{\sqrt{9+16}}\)
\(=\frac{5}{5}=1\)
\(\therefore\) Equation of the circle is
\(
(x-1)^{2}+(y-2)^{2}=1
\)
or \(\quad x^{2}+y^{2}-2 x-4 y+4=0\)
\(\Rightarrow \quad x-1=0 \text { and } y-2=0\)
\(\therefore \text {Radius}=\frac{3(1)+4(2)-6}{\sqrt{9+16}}\)
\(=\frac{5}{5}=1\)
\(\therefore\) Equation of the circle is
\(
(x-1)^{2}+(y-2)^{2}=1
\)
or \(\quad x^{2}+y^{2}-2 x-4 y+4=0\)
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