Write the standard equation of the circle with center ​(-1​,7​) that passes through the point ​(-1​,10​).?

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