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Let the mirror image of a circle c1 : x2 + y2 – 2x – 6y + a = 0 in line y = x + 1 be c2 : 5x2 + 5y2 + 10gx + 10fy + 38 = 0.  If r is the radius of circle c2, then a + 6r2 is equal to…………..

0 3 Views | Posted 2 months ago
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    Answered by

    Vishal Baghel | Contributor-Level 10

    2 months ago

    (x−1)2+(y−3)2=10−α

    x−y=−1x+y=a+b}→m(a+b−12,a+b+12)

    A’ = 2m – A = (b – 1, a + 1)

    C2:x2+y2+2gx+2fy+385=0

    r  g 2 + f 2 − c

    = 4 + 4 − 3 8 5 = 2 5

    C 1 : 2 5 = 1 + 9 − α = 1 0 − α

    α + 6 r 2 = 4 8 5 + 6 ( 2 5 ) = 6 0 5 = 1 2

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The distance between the two points A and A’ which lie on y = 2 such that both the line segments AB and A’B (where B is the point (2, 3)) subtend angle π4 at the origin, is equal to:
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Let A, A’ be (, 2) AB and A’B subtends π4 angle at (0, 0) slope of OA = 2α

 

slope of OB = 32

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⇒α+3α=± (4−3α2α)⇒α=10,  −25

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The equations of the sides AB, BC and CA of a triangle ABC are 2x + y = 0, x + py = 15a and x – y = 3 respectively. If its orthocenter is (2, a), −12<a<2, then p is equal to…………
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Vishal Baghel

Slope of AH = a+21 slope of BC = −1p∴p=a+2 ∴C (18p−30p+1, 15p−33p+1)

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A point P moves so that the sum of squares of its distances from the points (1, 2) and (2, 1) is 14. Let f(x, y) = 0 be the locus of P, which intersects the x-axis at the points A , B and the y-axis at the points C, D. Then the area of the quadrilateral ACBD is equal to:
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Vishal Baghel

Let point P : (h, k)

Therefore according to question,   (h−1)2+ (k−2)2+ (h+2)2+ (k−1)2=14

∴ locus of P (h, k) is x2+y2+x−3y−2=0

Now intersection with x – axis are x2+x−2=0⇒x=−2, 1

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Let equation of normal to x2 = y at Q (t, t2) is x + 2ty = t + 2t3

It passes through the point (1, -1) so, 2t3 + 3t – 1 = 0

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K-2h-11-23-1=-1⇒K=2h

? [? ABC]=55

⇒12 (5) (h-1)2+ (K-2)2=55

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