✔ ⋂ ⋃ ф ܓ ⋎ ∈ ≤ ≥ ∉ ⇒ → ∞ ⊙ ⊄ ⊂ ∴ ∵ ≠ ± ∀ ✕❌ △ ≌ ∠ ॥ π ₹ θ √ ⊥ 𝛂 β ܓ ⋎ λ ∃ ⇔🇽 RESOURCE CENTRE Lab Activities Mathematics 10+1 Activity-3 , Solution: The equation of the given circle is x^2 + y^2 + 8x - 16y + 64 = 0 ⇒ x 2 + 8 x + 16 + ( y 2 − 16 y + 64 ) = 16 \Rightarrow x^2 + 8x + 16 + (y^2 - 16y + 64) = 16 ⇒ ( x + 4 ) 2 + ( y − 8 ) 2 = 4 2 \Rightarrow (x + 4)^2 + (y - 8)^2 = 4^2 ⇒ { x − ( − 4 ) } 2 + ( y − 8 ) 2 = 4 2 \Rightarrow \{ x - (-4) \}^2 + (y - 8)^2 = 4^2 Clearly, the center of the circle is ( − 4 , 8 ) (-4, 8) ( − 4 , 8 ) and its radius is 4. The image of the center after reflection in the line x = 0 x = 0 x = 0 is ( 4 , 8 ) (4, 8) ( 4 , 8 ) . So, the equation of the reflected circle is ( x − 4 ) 2 + ( y − 8 ) 2 = 4 2 (x - 4)^2 + (y - 8)^2 = 4^2 Expanding the equation: x 2 − 8 x + y 2 − 16 y + 64 = 0 x^2 - 8x + y^2 - 16y + 64 = 0 Thus, the equation of the reflected circle is x 2 − 8 x + y 2 − 16 y + 64 = 0...
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✔ ⋂ ⋃ ф ܓ ⋎ ∈ ≤ ≥ ∉ ⇒ → ∞ ⊙ ⊄ ⊂ ∴ ∵ ≠ ± ∀ ✕❌ △ ≌ ∠ ॥ π ₹ θ √ ⊥ 𝛂 β ܓ ⋎ λ ∃ ⇔🇽 RESOURCE CENTRE Lab Activities Mathematics 10+1 Activity-3 , Solution: The equation of the given circle is x^2 + y^2 + 8x - 16y + 64 = 0 ⇒ x 2 + 8 x + 16 + ( y 2 − 16 y + 64 ) = 16 \Rightarrow x^2 + 8x + 16 + (y^2 - 16y + 64) = 16 ⇒ ( x + 4 ) 2 + ( y − 8 ) 2 = 4 2 \Rightarrow (x + 4)^2 + (y - 8)^2 = 4^2 ⇒ { x − ( − 4 ) } 2 + ( y − 8 ) 2 = 4 2 \Rightarrow \{ x - (-4) \}^2 + (y - 8)^2 = 4^2 Clearly, the center of the circle is ( − 4 , 8 ) (-4, 8) ( − 4 , 8 ) and its radius is 4. The image of the center after reflection in the line x = 0 x = 0 x = 0 is ( 4 , 8 ) (4, 8) ( 4 , 8 ) . So, the equation of the reflected circle is ( x − 4 ) 2 + ( y − 8 ) 2 = 4 2 (x - 4)^2 + (y - 8)^2 = 4^2 Expanding the equation: x 2 − 8 x + y 2 − 16 y + 64 = 0 x^2 - 8x + y^2 - 16y + 64 = 0 Thus, the equation of the reflected circle is x 2 − 8 x + y 2 − 16 y + 64 = 0...
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