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Question Number 69681 by ajfour last updated on 26/Sep/19

Commented by ajfour last updated on 26/Sep/19

$Q . 69496 (A t r y . .)$
Commented by ajfour last updated on 26/Sep/19

$R a d i u s o f c i r c l e .$
	Answered by ajfour last updated on 30/Sep/19

	Commented by ajfour last updated on 30/Sep/19
	$let cos α=a, cos β=h, cos γ=m, cos δ=u & sin α=b, sin β=k, sin γ=n, sin δ=v x_D =−qa+rm=ph−su ⇒ su+rm = qa+ph ...(i) y_D =qb+rn=pk+sv ⇒ sv−rn = qb−pk ...(ii) ∣CP ∣=R =∣CT∣ ⇒ _________________________ (−qa−m)^2 +(qb−n−R)^2 =R^2 [−qa+(r+6)m]^2 +(qb+(r+6)n−R]^2 =R^2 ⇒ r+6 and −1 are roots of eq. (−qa+mx)^2 +(qb+nx−R)^2 =R^2 (m^2 +n^2 )x^2 +2x[−qam+n(qb−R)] +q^2 a^2 +q^2 b^2 +R^2 −2qbR−R^2 =0 ⇒ x^2 +2x[−qam+n(qb−R)] +q^2 −2qbR=0 _________________________ ⇒ r+5=qam−n(qb−R) & m^2 +n^2 =1 .....(iii),(iv) [r+5+n(qb−R)]^2 =q^2 a^2 (1−n^2 ) ⇒ n^2 {(qb−R)^2 +q^2 a^2 }+ +2(r+5)(qb−R)n+(r+5)^2 −q^2 a^2 =0 _________________________ ∣CM∣=R =∣CS∣ ⇒ (ph+3u)^2 +(pk−3v−R)^2 =R^2 [ph−(s+5)u]^2 +[pk+(s+5)v−R]^2 =R^2 ⇒ s+5 and −3 are roots of eq. (ph−ux)^2 +(pk+vx−R)^2 =R^2 _________________________ ⇒ sum of roots (s+5)+(−3)= s+2=phu−v(pk−R) (as h^2 +k^2 =1) & u^2 +v^2 =1 ......(v),(vi) _________________________ plus _________________________ a^2 +b^2 =1 , h^2 +k^2 =1 ..(vii)&(viii) 2q=r+6 ....(ix) 4p=3(s+5) ....(x) 5(s+3)=6(r+1) ....(xi) further 2bR=q+2 & ..(xii) 2kR=p+4 ....(xiii) Thirteen eqs. in thirteen unknowns a,b,h,k,m,n,u,v,p,q,r,s,&R.$
	$l e t \cos α = a, \cos β = h, \cos γ = m,$ $\cos δ = u & \sin α = b, \sin β = k,$ $\sin γ = n, \sin δ = v$ $x_{D} = - q a + r m = p h - s u$ $\Rightarrow s u + r m = q a + p h . . . (i)$ $y_{D} = q b + r n = p k + s v$ $\Rightarrow s v - r n = q b - p k . . . (i i)$ $∣ C P ∣= R =∣ C T ∣ \Rightarrow$ $_________________________$ ${(- q a - m)}^{2} + {(q b - n - R)}^{2} = R^{2}$ ${[- q a + (r + 6) m]}^{2} + {(q b + (r + 6) n - R]}^{2} = R^{2}$ $\Rightarrow r + 6 a n d - 1 a r e r o o t s o f e q .$ ${(- q a + m x)}^{2} + {(q b + n x - R)}^{2} = R^{2}$ $(m^{2} + n^{2}) x^{2} + 2 x [- q a m + n (q b - R)]$ $+ q^{2} a^{2} + q^{2} b^{2} + R^{2} - 2 q b R - R^{2} = 0$ $\Rightarrow x^{2} + 2 x [- q a m + n (q b - R)]$ $+ q^{2} - 2 q b R = 0$ $_________________________$ $\Rightarrow r + 5 = q a m - n (q b - R)$ $& m^{2} + n^{2} = 1 . . . . . (i i i), (i v)$ ${[r + 5 + n (q b - R)]}^{2} = q^{2} a^{2} (1 - n^{2})$ $\Rightarrow n^{2} {{(q b - R)}^{2} + q^{2} a^{2}} +$ $+ 2 (r + 5) (q b - R) n + {(r + 5)}^{2} - q^{2} a^{2} = 0$ $_________________________$ $∣ C M ∣= R =∣ C S ∣ \Rightarrow$ ${(p h + 3 u)}^{2} + {(p k - 3 v - R)}^{2} = R^{2}$ ${[p h - (s + 5) u]}^{2} + {[p k + (s + 5) v - R]}^{2} = R^{2}$ $\Rightarrow s + 5 a n d - 3 a r e r o o t s o f e q .$ ${(p h - u x)}^{2} + {(p k + v x - R)}^{2} = R^{2}$ $_________________________$ $\Rightarrow s u m o f r o o t s (s + 5) + (- 3) =$ $s + 2 = p h u - v (p k - R) (a s h^{2} + k^{2} = 1)$ $& u^{2} + v^{2} = 1 . . . . . . (v), (v i)$ $_________________________$ $p l u s$ $_________________________$ $a^{2} + b^{2} = 1, h^{2} + k^{2} = 1 . . (v i i) & (v i i i)$ $2 q = r + 6 . . . . (i x)$ $4 p = 3 (s + 5) . . . . (x)$ $5 (s + 3) = 6 (r + 1) . . . . (x i)$ $f u r t h e r 2 b R = q + 2 & . . (x i i)$ $2 k R = p + 4 . . . . (x i i i)$ $T h i r t e e n e q s . i n t h i r t e e n u n k n o w n s$ $a, b, h, k, m, n, u, v, p, q, r, s, & R .$
	Answered by mr W last updated on 30/Sep/19

	Commented by mr W last updated on 30/Sep/19

	${(p + 4)}^{2} = 4 (R^{2} - a^{2})$ $\Rightarrow 4 a^{2} = 4 R^{2} - {(p + 4)}^{2}$ ${(q + 2)}^{2} = 4 (R^{2} - b^{2})$ $\Rightarrow 4 b^{2} = 4 R^{2} - {(q + 2)}^{2}$ ${(r + 6 + 1)}^{2} = 4 (R^{2} - c^{2})$ $\Rightarrow 4 c^{2} = 4 R^{2} - {(r + 7)}^{2}$ ${(s + 5 + 3)}^{2} = 4 (R^{2} - d^{2})$ $\Rightarrow 4 d^{2} = 4 R^{2} - {(s + 8)}^{2}$ $a^{2} + {(p - \frac{p + 4}{2})}^{2} = b^{2} + {(q - \frac{q + 2}{2})}^{2}$ $4 a^{2} + {(p - 4)}^{2} = 4 b^{2} + {(q - 2)}^{2}$ $4 R^{2} - {(p + 4)}^{2} + {(p - 4)}^{2} = 4 R^{2} - {(q + 2)}^{2} + {(q - 2)}^{2}$ $\Rightarrow 2 p = q$ $b^{2} + {(\frac{q + 2}{2} - 2)}^{2} = c^{2} + {(\frac{r + 6 + 1}{2} - 1)}^{2}$ $4 b^{2} + {(q - 2)}^{2} = 4 c^{2} + {(r + 5)}^{2}$ $4 R^{2} - {(q + 2)}^{2} + {(q - 2)}^{2} = 4 R^{2} - {(r + 7)}^{2} + {(r + 5)}^{2}$ $\Rightarrow 2 q = r + 6$ $c^{2} + {(\frac{r + 6 + 1}{2} - 6)}^{2} = d^{2} + {(\frac{s + 5 + 3}{2} - 5)}^{2}$ $4 c^{2} + {(r - 5)}^{2} = 4 d^{2} + {(s - 2)}^{2}$ $4 R^{2} - {(r + 7)}^{2} + {(r - 5)}^{2} = 4 R^{2} - {(s + 8)}^{2} + {(s - 2)}^{2}$ $\Rightarrow 6 (r + 1) = 5 (s + 3)$ $d^{2} + {(\frac{s + 5 + 3}{2} - 3)}^{2} = a^{2} + {(\frac{p + 4}{2} - 4)}^{2}$ $4 d^{2} + {(s + 2)}^{2} = 4 a^{2} + {(p - 4)}^{2}$ $4 R^{2} - {(s + 8)}^{2} + {(s + 2)}^{2} = 4 R^{2} - {(p + 4)}^{2} + {(p - 4)}^{2}$ $\Rightarrow 3 (s + 5) = 4 p$ $q = 2 p$ $r = 2 q - 6 = 4 p - 6$ $s = \frac{6}{5} (r + 1) - 3 = \frac{24 p}{5} - 9$ $3 (\frac{24 p}{5} - 9) = 4 p$ $3 (24 p - 45) = 20 p$ $\Rightarrow p = \frac{135}{52}$ $\Rightarrow q = \frac{135}{26}$ $\Rightarrow r = 4 \times \frac{135}{52} - 6 = \frac{57}{13}$ $\Rightarrow s = \frac{24}{5} \times \frac{135}{52} - 9 = \frac{45}{13}$ $. . . . . .$
	Commented by mr W last updated on 30/Sep/19

	$w e c a n a l s o g e t d i r e c t l y :$ $5 (s + 3) = 6 (r + 1)$ $4 p = 3 (s + 5)$ $2 q = 1 \times (r + 6)$
	Commented by ajfour last updated on 01/Oct/19

	$T h a n k s S i r, c a n w e n o w$ $c a l c u l a t e R a d i u s ?$
	Commented by mr W last updated on 02/Oct/19

	$t h e r a d i u s s h o u l d b e u n i q u e, b u t i$ $h a v e n o t g o t a w a y t o c a l c u l a t e .$
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