If P is any point of C 1 and Q is another point on C 2 , then 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 [2006]

Question

$A B C D$ is a square of side length 2 units. $C_{1}$ is the circle touching all the sides of the square $A B C D$ and $C_{2}$ is the circumcircle of square $A B C D$ . $L$ is a fixed line in the same plane. [2006]

Q. If $P$ is any point of $C_{1}$ and $Q$ is another 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

Smart Achievers · Accepted Answer

(1)

According to the given question, we can assume the square $A B C D$ with its vertices $A (1, 1), B (- 1, 1), C (- 1, - 1), D (1, - 1) .$

$P be the point (0, 1) and Q be the point (\sqrt{2}, 0) .$

Then, $\frac{P A^{2} + P B^{2} + P C^{2} + P D^{2}}{Q A^{2} + Q B^{2} + Q C^{2} + Q D^{2}}$

$= \frac{1 + 1 + 5 + 5}{2 [{(\sqrt{2} - 1)}^{2} + 1] + 2 [{(\sqrt{2} + 1)}^{2} + 1]}$

$= \frac{12}{16} = 0.75$

Solution

1 0.75

2 1.25

3 1

4 0.5

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Solution

Q. ABCD is a square of side length 2 units. C1 is the circle touching all the sides of the square ABCD and C2 is the circumcircle of square ABCD. L is a fixed line in the same plane. [2006] Q. If P is any point of C1 and Q is another point on C2, then PA2+PB2+PC2+PD2QA2+QB2+QC2+QD2 is equal to

1 0.75

2 1.25

3 1

4 0.5

Ans. (1) According to the given question, we can assume the square ABCD with its vertices A(1,1), B(-1,1), C(-1,-1), D(1,-1). P be the point (0,1) and Q be the point (2,0). Then, PA2+PB2+PC2+PD2QA2+QB2+QC2+QD2 =1+1+5+52[(2-1)2+1]+2[(2+1)2+1] =1216=0.75

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