Solving
1. Finding the value of a:
Chunki PR diameter hai, toh circle par kisi bhi point Q ke liye angle ∠PQR=90∘ hoga.
Iska matlab slope of PQ× slope of QR=−1
(9−(−3)10−2)×(9−a10−4)=−1
(128)×(9−a6)=−1⟹(32)×(9−a6)=−1
Toh point R ke coordinates (13,4) hain.
2. Equation of Circle C:
Diameter form ka use karke (x−x1)(x−x2)+(y−y1)(y−y2)=0:
x2−10x−39+y2−6y+8=0⟹x2+y2−10x−6y−31=0
3. Finding Point S (Intersection of tangents at Q and R):
Point S(h,l) ke liye QR chord of contact hai. Equation of chord of contact: T=0
x(h−5)+y(l−3)−(5h+3l+31)=0
Yeh line aur QR ki equation same honi chahiye. QR ki equation points (9,10) aur (13,4) se:
y−4=9−1310−4(x−13)⟹y−4=−46(x−13)⟹2y−8=−3x+39
Ab coefficients compare karte hain:
Pehle do parts se: 2h−10=3l−9⟹2h−3l=1
Solve karne par humein intersection point S ke coordinates milte hain: S(7,313)
4. Finding k:
Point S(7,313) line 2x−ky=1 par lie karta hai:
Answer: k=3
