AP EAMCET · Maths · Circle
A circle touches the line \(2 x+y-10=0\) at \((3,4)\) and passes through the point \((1,-2)\). Then a point that lies on the circle is
- A (5,4)
- B (4,5)
- C (-5,4)
- D (4,-5)
Answer & Solution
Correct Answer
(C) (-5,4)
Step-by-step Solution
Detailed explanation
\((x-3)^2+(y-4)^2+\lambda(2x+y-10)=0\) \((1-3)^2+(-2-4)^2+\lambda(2(1)-2-10)=0\) \(4+36+\lambda(-10)=0\) \(40-10\lambda=0 \implies \lambda=4\) \((x-3)^2+(y-4)^2+4(2x+y-10)=0\) \(x^2-6x+9+y^2-8y+16+8x+4y-40=0\) \(x^2+y^2+2x-4y-15=0\) Test point \((-5,4)\):…
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