SATHEE: Circle 2 Question 17

Circle 2 Question 17

18. If $P$ is a point of $C_{1}$ and $Q$ is a point on $C_{2}$ , then $\frac{P A^{2} + P B^{2} + P C^{2} + P D^{2}}{Q A^{2} + Q B^{2} + Q C^{2} + Q D^{2}}$ is equal to

(a) 0.75

(b) 1.25

(d) 0.5

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Solution:

Let the, equation of circles are

$C_{1} : (x - 1)^{2} + (y - 1)^{2} = (1)^{2}$

and $C_{2} : (x - 1)^{2} + (y - 1)^{2} = (\sqrt{2})^{2}$

$∴$ Coordinates of $P (1 + \cos θ, 1 + \sin θ)$

$\begin{aligned} and Q (1 + \sqrt{2} \cos θ, 1 + \sqrt{2} \sin θ) \\ ∴ P A^{2} + P B^{2} + P C^{2} + P D^{2} \\ = (1 + \cos θ)^{2} + (1 + \sin θ)^{2} + (\cos θ - 1)^{2} + (1 + \sin θ)^{2} \\ + (\cos θ - 1)^{2} + (\sin θ - 1)^{2} \\ + (1 + \cos θ)^{2} + (\sin θ - 1)^{2} = 12 \end{aligned}$

Similarly, $Q A^{2} + Q B^{2} + Q C^{2} + Q D^{2} = 16$

$∴ \frac{Σ P A^{2}}{Σ Q A^{2}} = \frac{12}{16} = 0.75$

Circle 2 Question 17

18. If $P$ is a point of $C_{1}$ and $Q$ is a point on $C_{2}$ , then $\frac{P A^{2} + P B^{2} + P C^{2} + P D^{2}}{Q A^{2} + Q B^{2} + Q C^{2} + Q D^{2}}$ is equal to

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Circle 2 Question 17

18. If P is a point of C1 and Q is a point on C2, then PA2+PB2+PC2+PD2QA2+QB2+QC2+QD2 is equal to

Solution:

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18. If $P$ is a point of $C_{1}$ and $Q$ is a point on $C_{2}$ , then $\frac{P A^{2} + P B^{2} + P C^{2} + P D^{2}}{Q A^{2} + Q B^{2} + Q C^{2} + Q D^{2}}$ is equal to