KEAM 2010 Math Paper

∵(4,7) and (−2,−1) are end points of circle.
⇒ Centre of circle
=(‌

4−2

2

,‌

7−1

2

)=(1,3)
∴ Slope of AC=‌

−1−0

−2−a

=−‌

1

2+a

and slope of AD=‌

7−0

4−a

=‌

7

4−a

since, ∆CAD is right angle as CD is diameter.
⇒‌‌ (slope of AC (slope of AD)=−1
⇒‌‌(‌

1

2+a

)(‌

7

4−a

)=−1
⇒‌‌‌

7

−a2+2a+8

=−1
⇒‌‌a2−2a−15=0
⇒‌‌a=−3,5
Similarly, ‌‌b=−3,5
∴ Point A=(−3,0) and piont B=(5,0)
∴‌‌AB=√(−3−5)2+(0−0)2
=√(−8)2=8