Write the standard equation of the circle with center (-1,7) that passes through the point (-1,10).?
回答 (5)
Radius of the circle, r
= |10 - 7|
= 3
Hence, r² = 9
The standard equation of the circles with center (-1, 7) is
(x + 1)² + (y - 7)² = 9
Circle, center at (-1,7), passes through P(-1,10).;
Eqn of circle is (x+1)^2 + (y-7)^2 = r^2....(1). P is on (1). Therefore;
(-1+1)^2 + (10-7)^2 = r^2, ie., r^2 = 9 & eqn of circle is (x+1)^2 + (y-7)^2 = 9.
It should be apparent that the radius is three - look at the points again. That and the centre give you the information for the equation.
Otherwise, the equation of the circle centred at (x₀,y₀) and passing through (x₁,y₁) is (x-x₀)²+(y-y₀)² = (x₁-x₀)²+(y₁-y₀)².
Here this is (x - -1)² + (y - 7)² = (-1 - -1)² + (10 - 7)², which simplifies to (x + 1)² + (y - 7)² = 9.
If ' r ' is radius of a circle, ( h, k ) are coordinates of its center, its equation is -
( x - h )² + ( y - k )² = r²
=> [ x - (-1) ]² + [ y - (7) ]² = r²
=> ( x + 1 )² + ( y - 7 )² = r² ..................(1)
We have, now, to find out the value of Radius using the other condition (passes
through the point (-1,10).
Since the circle : ( x + 1 )² + ( y - 7 )² = r² passes through ( - 1, 10 ), these
coordinates must satisfy the eqn.
=> ( - 1 + 1 )² + ( 10 - 7 )² = r²
=> r² = 3² = 9
Hence the required equation is -----
( x + 1 )² + ( y - 7 )² = 9 ...................... Answer
(x - (-1))^2 + (y - 7)^2 = r^2
(x + 1)^2 + (y - 7)^2 = r^2
(-1 + 1)^2 + (10 - 7)^2 = r^2
0^2 + 3^2 = r^2
9 = r^2
(x + 1)^2 + (y - 7)^2 = 9
收錄日期: 2021-04-18 18:36:51
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200626231745AAc9cFU
檢視 Wayback Machine 備份