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Chapter 1: Chapter 1

Problem 4

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Find the equation of the circle passing through the points of intersection of \(x^{2}+y^{2}=6\) and \(x^{2}+y^{2}-6 x+8=0\) and the point \((1,1)\).

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The equation of the circle passing through the intersection points and the point \((1,1)\) is \((x - 4/3)^{2} + y^{2} = 4\).
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TABLE OF CONTENTS :
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Step 1: Find the intersection points

Subtract the second equation from the first to find the points of intersection. Which gives \(6x - 8 = 0\), solving this we get \(x = 4/3\). Substitute \(x = 4/3\) in the first equation, we get \(y = \pm \sqrt{6 - (4/3)^2}\), which yields two points of intersection: \(A(4/3, \sqrt{14}/3)\) and \(B(4/3, -\sqrt{14}/3)\).

Step 2: Form the system of equations

The standard equation of a circle is \((x-a)^{2} + (y-b)^{2} = r^{2}\), where \((a, b)\) is the center and \(r\) is the radius. Substitute point A, B and the given point \((1,1)\) into this equation, we will get three equations.

Step 3: Solve the system of equations

Solving these equations simultaneously, will give the values \(a = 4/3\), \(b = 0\), and \(r = 2\). Therefore, the equation of the desired circle is \((x - 4/3)^{2} + y^{2} = 4\).

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Most popular questions from this chapter

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The equation of the circle which touches the axes of co-ordinates and the line \(\frac{x}{3}+\frac{y}{4}=1\) and whose centres lies in the first quadrant is $$ x^{2}+y^{2}-2 \lambda x-2 \lambda y+\lambda^{2}=0 $$ which \(\lambda\) is equal to: (a) 1 (b) 2 (c) 3 (d) 6
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The equation of the image of the circle \((x-3)^{2}+(y-2)^{2}=1\) by the line mirror \(x+y=19\) is: (a) \((x-14)^{2}+(y-13)^{2}=1\) (b) \((x-15)^{2}+(y-14)^{2}=1\) (c) \((x-16)^{2}+(y-15)^{2}=1\) (d) \((x-17)^{2}+(y-16)^{2}=1\)
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If the circle \(x^{2}+y^{2}+2 g x+2 f y+c=0\) cuts each of the circles \(x^{2}+y^{2}-4=0\), \(x^{2}+y^{2}-6 x-8 y+10=0\) and \(x^{2}+y^{2}+2 x-4 y-2=0\) at the extremities of a diameter, then : (b) \(g+f=c-1\) (a) \(c=-4\) (d) \(g f=6\) (c) \(g^{2}+f^{2}-c=17\)
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If a chord of the circle \(x^{2}+y^{2}-4 x-2 y-c=0\) is trisected at the poing \((1 / 3,1 / 3)\) and \((8 / 3,8 / 3)\) then : (a) \(c=10\) (b) \(c=20\) (c) \(c=15\) (d) \(c^{2}-40 c+400=0\)
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If \((a, b)\) is a point on the chord \(A B\) of the circle, where the ends of the chord are \(A=(-2,3)\) and \(B \equiv(3,2)\) then : (a) \(a \in[-3,2], b \in[2,3]\) (b) \(a \in[2,3], b \in[-3,2]\) (c) \(a \in[-2,2], b \in[-3,3]\) (d) \(a \in[-3,3], b \in[-2,2]\)
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