JEE Main & Advanced JEE Main Paper (Held On 15 April 2018) Slot-I

  • question_answer
    A circle passes through the points \[(2,3)\] and \[(4,5)\]. If its centre lies on the line, \[y-4x+3=0\], then its radius is equal to                                                                                            [JEE Online 15-04-2018]

    A) \[\sqrt{5}\]                              

    B) \[1\]   

    C) \[\sqrt{2}\]                              

    D) \[2\]

    Correct Answer: D

    Solution :

    Equation of the line through the given points is \[y-3=x-2\Rightarrow x-y+1=0\] Equation of the perpendicular line through the midpoint \[(3,4)\] is\[x+y-7=0\]. This intersects the given line at the center of the circle. So, the center of the circle is found to be \[(2,5)\]. Clearly, the radius is then \[\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}}=\sqrt{{{(2-2)}^{2}}+{{(3-5)}^{2}}}=2\]units . So the answer is option D.


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