Circle 4 Question 5

5. The circle passing through $(1,-2)$ and touching the axis of $x$ at $(3,0)$ also passes through the point

(2013 Main)

(a) $(-5,2)$

(b) $(2,-5)$

(c) $(5,-2)$

(d) $(-2,5)$

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Answer:

Correct Answer: 5. (c)

Solution:

  1. Let the equation of circle be $(x-3)^{2}+(y-0)^{2}+\lambda y=0$

As it passes through $(1,-2)$

$$ \begin{array}{ll} \therefore & (1-3)^{2}+(-2)^{2}+\lambda(-2)=0 \\ \Rightarrow & 4+4-2 \lambda=0 \Rightarrow \lambda=4 \end{array} $$

$\therefore$ Equation of circle is

$$ (x-3)^{2}+y^{2}+4 y=0 $$

By hit and trial method, we see that point $(5,-2)$ satisfies equation of circle.



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