The radius of a circle with centre (-2,3) is 5 units .the point (2,5) lies
Answers
Answer:
lies inside the circle
Explanation:
given center of the circle is (-2,3) and radius 5
the circle equation having center (h,k) is
the circle equation having center (h,k) is(x-h)^2+(y-k)^2=r^2
here (h,k)= (-2,3) and r=5
then the equation of circle with centre (-2,3) and radius 5 is
(x-(-2))^2+(y-3)^2=(5)^2
(x+2)^2+(y-3)^2=25
x^2+4+4x+y^2+9-6y=25
x^2+y^2+4x-6y+13=25
x^2+y^2+4x-6y-12=0
x^2+y^2+4x-6y-12=0 this is equation of circle of given center and radius
concept:
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if point (x1,y1) lies
(i) lies on the circle, if when we substitute the
point in circle equation then the value should
be equal to zero
(ii) lies inside the circle , if when when we
substitute the point in circle equation then the
value should be less than zero
(iii) lies outside the circle, if when when we
substitute the point in circle equation then the
value should be greater than zero
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based on the conditions we have to consider where the point lies
here (x1,y1) is (2,5)
substitute the point in circle equation
= (2)^2+(5)^2+4(2)-6(5)-12
= 4+25+8-30-12
= -5<0
here the value is less than zero
therefore , the given point lies inside the circle