SATHEE: Circle 4 Question 12

Circle 4 Question 12

12. Two circles $x^{2} + y^{2} = 6$ and $x^{2} + y^{2} - 6 x + 8 = 0$ are given. Then the equation of the circle through their points of intersection and the point $(1, 1)$ is

$(1980, 1 M)$

(a) $x^{2} + y^{2} - 6 x + 4 = 0$

(b) $x^{2} + y^{2} - 3 x + 1 = 0$

(d) None of these

Show Answer

Answer:

Correct Answer: 12. (b)

Solution:

The required equation of circle is, $S_{1} + λ (S_{2} - S_{1}) = 0$ .

$∴ (x^{2} + y^{2} - 6) + λ (- 6 x + 14) = 0$

Also, passing through $(1, 1)$ .

$\begin{aligned} \Rightarrow & - 4 + λ (8) & = 0 \\ \Rightarrow & λ & = \frac{1}{2} \end{aligned}$

$∴$ Required equation of circle is

$or \begin{aligned} x^{2} + y^{2} - 6 - 3 x + 7 & = 0 \\ x^{2} + y^{2} - 3 x + 1 & = 0 \end{aligned}$

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Circle 4 Question 12

12. Two circles x2+y2=6 and x2+y2−6x+8=0 are given. Then the equation of the circle through their points of intersection and the point (1,1) is

Answer:

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12. Two circles $x^{2} + y^{2} = 6$ and $x^{2} + y^{2} - 6 x + 8 = 0$ are given. Then the equation of the circle through their points of intersection and the point $(1, 1)$ is