jarak titik a(7,2) kelingkaran x^2+y^2-10x+4y-151=0 adalah tolong dibantuya

Persamaan Lingkaran
= x²+y²-10x+4y-151 = 0

A = -10 B = 4 C = -151

Titik Pusat = (-½A, -½B) = (5, -2)

Jari-jari (r)
[tex] r = \sqrt{ \frac{1}{4} {A}^{2} + \frac{1}{4} {B}^{2} - C} \\ = \sqrt{25 + 4 - ( - 151)} = \sqrt{180} \\ = \sqrt{4 \times 9 \times 5} = 2 \times 3 \times \sqrt{5 } \\ = 6 \sqrt{5} [/tex]

a(7,2) dan pusat (5,-2)

[tex]jarak = \sqrt{{(x2-x1)}^{2} + {(y2-y1)}^{2} } \\ = \sqrt{{(7-5)}^{2} + {(2 - ( - 2))}^{2}} \\ = \sqrt{{2}^{2} + {4}^{2}} = \sqrt{4 + 16 } = \sqrt{20} \\ = 2 \sqrt{5} [/tex]
jawabannya (a) 2√5