If A,B are the points of contact of the tangents drawn from the point (−3,1) to the circle x2+y2−4x+2y−4=0, then the equation of the circumcircle of the △PAB is
The equation of the circle is 44 x2+y2−4x+2y−4=0 Comparing this with the general form x2+y2+2gx+2fy+c=0 We have ‌2g‌=−4 ⇒‌g‌=−2‌ and ‌2f=2 ⇒‌f‌=1 So, the center of the circle C is (−g,−f)=(2,−1) And, the radius is r=√g2+f2−c ‌=√(−2)2+(1)2−(−4) ‌=√4+1+4=√9=3 The point from which tangents are drawn is P=(−3,1) Now, the circumcircle of △PAB is the circle with PC as its diameter. The center of this circumcircle is the mid-point of PC. Mid-point M=(‌
−3+2
2
,‌
1+(−1)
2
) ⇒(‌
−1
2
,‌
0
2
)=(‌
−1
2
,0) Distance PC=√(2−(−3))2+(−1−1)2 ‌⇒‌‌√52+(−2)2 ‌⇒‌‌√25+4=√29 ∴ The radius of the circumcircle =‌
1
2
⋅‌ distance ‌PC=‌
√29
2
Now, the equation of a circle with center ( h,k ) and radius R is (x−h)2+(y−k)2=R2 Here, (h,k)=(‌