Did you forget to get the square root of the denominator; numerator shouldn't even be squared? Equation should be Ah + Bk + C / the square root of A squared + B squared.
Find the equation of the circle with the center at (-4,2) and tangent line to the line 2x-y+2=0 h= -4 k= 2 2x-y+2=0 A= 2 B= 1 C= 2 r² (Ah+Bk+C)² over A²+B² = (2(-4)+(1)(2)+2)² over 2²+1 = (-8+2+1)² over 4+1 = (-11)² over 5 = 125 over 5 => r² = 24.2 (x+4)²+(y-2)² = 24.2
how about the center is not given then the tangent line is x+y=2?????
Give me the problem?
what if only 0ne center? and given the center of tangent?
Whats the problem?
Did you forget to get the square root of the denominator; numerator shouldn't even be squared? Equation should be Ah + Bk + C / the square root of A squared + B squared.
The answer should be negative square root of 13; or -3.61
Hello…i am using r² formula so that easy to find the value… that formula you said if you’re solving for r.
hello sir, may i ask if my answer is right
Find the equation of the circle with the center at (-4,2) and tangent line to the line 2x-y+2=0
h= -4
k= 2
2x-y+2=0
A= 2
B= 1
C= 2
r² (Ah+Bk+C)² over A²+B²
= (2(-4)+(1)(2)+2)² over 2²+1
= (-8+2+1)² over 4+1
= (-11)² over 5
= 125 over 5 => r² = 24.2
(x+4)²+(y-2)² = 24.2
paki correct po ako hehe needed lang po
Whats problem?
hello sir, may you check sir if my answer was right hehehe, nasa taas po
Given : Find the equation of the circle with the center at (-4,2) and tangent line to the line 2x-y+2=0