MHT CET 2021 Math Solved Paper

Given, equation of circles are x2+y2−4y=0 and x2+y2−8x−4y+11=0
∴ Equation of chord is
x2+y2−4y−(x2+y2−8x−4y+11)=0
⇒‌‌8x−11=0 So, centre and radius of first circle are O(0,2) and OP=r=2.
Now, perpendicular distance from O(0,2) to the line 8x−11 is
d=OM=‌

|8×0−11|

√82

=‌

11

8

‌ In ‌∆OMP,PM‌‌=√OP2−OM2
‌‌=√22−(‌

11

8

)2=√4−‌

121

64

‌‌=√‌

256−121

64

=‌

√135

8

∴‌ Length of chord ‌PQ=2PM
=2×‌

√135

8

=‌

√135

4

cm