Let circle C1: x2+y-42=12 intersect circle C2: x-32+y2=13 at A and B A quadrilateral ADBC is formed by the - Maths - Straight Lines

Question

Let circle C 1 :   x 2 + y - 4 2 = 12   i n t e r s e c t
  c i r c l e   C 2 :   x - 3 2 + y 2 = 13 at A and
B A quadrilateral ADBC is formed by the tangents at A and B to both
the circles The diameter of the circum-circle of quadrilateral ADBC
is

(A) 4 (B) 5 (C) 6 (D) 9 25

Shivam Mishra · Accepted Answer

Dear Student,
Please find below the solution to the asked query:

We have C1:x2+y-42=12C2:x-32+y2=13Equation of common chord AB is C1-C2=0x2+y-42-x-32-y2=12-13⇒x2+y2+16-8y-x2-9+6x-y2+1=0⇒6x-8y+8=0⇒3x-4y+4=0
iNow tangents at A and B to C1 intersect at C, hence with respect to CAB will be chord of contact with respect to C
C1:x2+y2+16-8y=12C1::x2+y2-8y+4=0Let C be h,kchord of contact is T=0⇒xh+yk-8y+k2+4=0⇒xh+yk-4y-4k+4=⇒xh+yk-4+4-4k=0
iii and ii represent same equationh3=k-4-4=4-4k4k-4-4=4-4k4⇒k-4=4k-4⇒k=0h3=k-4-4h3=0-4-4h=3Hence C is 3,0Unique Circle passes through A, B and C , hence it will automaticallypass through D as per question
We know that circle passing through A, B and C also passes through cenof C10,4 and C and C1 are diametric
HenceDiameter=CC1=3-02+0-42=5

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Let circle C 1 : x 2 + y - 4 2 = 12 i n t e r s e c t c i r c l e C 2 : x - 3 2 + y 2 = 13 at A and B. A quadrilateral ADBC is formed by the tangents at A and B to both the circles. The diameter of the circum-circle of quadrilateral ADBC is (A) 4 (B) 5 (C) 6 (D) 9.25

Let circle $C_{1} : x^{2} + {(y - 4)}^{2} = 12 i n t e r s e c t c i r c l e C_{2} : {(x - 3)}^{2} + y^{2} = 13$ at A and B. A quadrilateral ADBC is formed by the tangents at A and B to both the circles. The diameter of the circum-circle of quadrilateral ADBC is

(A) 4 (B) 5 (C) 6 (D) 9.25