NIMCET 2025 — Mathematics PYQ
NIMCET | Mathematics | 2025The circle is the image of the circle across the line . The value of is:
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The circle x2+y2+αx+βy+γ=0 is the image of the circle x2+y2−6x−10y+30=0 across the line 3x+y=2. The value of ∣α+β+γ∣ is:
23.6
(Correct Answer)22
20
21
23.6
The given equation of the original circle is:
x2+y2−6x−10y+30=0
Comparing this with the standard general equation of a circle x2+y2+2gx+2fy+c=0:
2g=−6⟹g=−3
2f=−10⟹f=−5
c=30
Center (C1): (−g,−f)=(3,5)
Radius (r): g2+f2−c=(−3)2+(−5)2−30=9+25−30=4=2
Let the center of the new image circle be C2(h,k). This point C2 is the reflection of the point C1(3,5) across the line mirror 3x+y−2=0.
Using the standard point-reflection formula for a point (x1,y1) about a line ax+by+d=0:
ah−x1=bk−y1=−2(a2+b2ax1+by1+d)
Substituting x1=3, y1=5, a=3, b=1, and d=−2:
3h−3=1k−5=−2(32+123(3)+1(5)−2)
3h−3=1k−5=−2(109+5−2)
3h−3=1k−5=−2(1012)=−2.4
Now, solving individual components for h and k:
For h:
3h−3=−2.4⟹h−3=−7.2⟹h=−4.2
For k:
1k−5=−2.4⟹k=5−2.4⟹k=2.6
Therefore, the center of the reflected image circle is C2(−4.2,2.6).
Since the radius remains identical (r=2), we use the central form of the circle equation with center (h,k)=(−4.2,2.6):
(x−h)2+(y−k)2=r2
(x+4.2)2+(y−2.6)2=22
Expanding the squares:
(x2+8.4x+17.64)+(y2−5.2y+6.76)=4
x2+y2+8.4x−5.2y+(17.64+6.76−4)=0
x2+y2+8.4x−5.2y+20.4=0
Comparing our expanded equation with the target format x2+y2+αx+βy+γ=0:
α=8.4
β=−5.2
γ=20.4
α+β+γ=8.4+(−5.2)+20.4
α+β+γ=3.2+20.4=23.6
Taking the absolute value:
∣α+β+γ∣=∣23.6∣=23.6
Final Answer: 23.6
The given equation of the original circle is:
x2+y2−6x−10y+30=0
Comparing this with the standard general equation of a circle x2+y2+2gx+2fy+c=0:
2g=−6⟹g=−3
2f=−10⟹f=−5
c=30
Center (C1): (−g,−f)=(3,5)
Radius (r): g2+f2−c=(−3)2+(−5)2−30=9+25−30=4=2
Let the center of the new image circle be C2(h,k). This point C2 is the reflection of the point C1(3,5) across the line mirror 3x+y−2=0.
Using the standard point-reflection formula for a point (x1,y1) about a line ax+by+d=0:
ah−x1=bk−y1=−2(a2+b2ax1+by1+d)
Substituting x1=3, y1=5, a=3, b=1, and d=−2:
3h−3=1k−5=−2(32+123(3)+1(5)−2)
3h−3=1k−5=−2(109+5−2)
3h−3=1k−5=−2(1012)=−2.4
Now, solving individual components for h and k:
For h:
3h−3=−2.4⟹h−3=−7.2⟹h=−4.2
For k:
1k−5=−2.4⟹k=5−2.4⟹k=2.6
Therefore, the center of the reflected image circle is C2(−4.2,2.6).
Since the radius remains identical (r=2), we use the central form of the circle equation with center (h,k)=(−4.2,2.6):
(x−h)2+(y−k)2=r2
(x+4.2)2+(y−2.6)2=22
Expanding the squares:
(x2+8.4x+17.64)+(y2−5.2y+6.76)=4
x2+y2+8.4x−5.2y+(17.64+6.76−4)=0
x2+y2+8.4x−5.2y+20.4=0
Comparing our expanded equation with the target format x2+y2+αx+βy+γ=0:
α=8.4
β=−5.2
γ=20.4
α+β+γ=8.4+(−5.2)+20.4
α+β+γ=3.2+20.4=23.6
Taking the absolute value:
∣α+β+γ∣=∣23.6∣=23.6
Final Answer: 23.6