If you want identical parameters, you should just pool your data:
df2 <- data.frame(Y=c(df$B,df$C), X=rep(df$A, 2))
p <- nls(Y ~ k1/X + k2,
data = df2,
start = list(k1=10, k2=10),
lower = c(0, 0),
algorithm = "port")
summary(p)
# Formula: Y ~ k1/X + k2
#
# Parameters:
# Estimate Std. Error t value Pr(>|t|)
# k1 124.819 18.078 6.904 0.000124 ***
# k2 1.199 9.781 0.123 0.905439
# ---
# Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
#
# Residual standard error: 16.59 on 8 degrees of freedom
#
# Algorithm "port", convergence message: both X-convergence and relative convergence (5)
Edit:
If you want one parameter to be equal and one to vary, you could use a mixed effects model. However, I don't know how to specify constraints for that (I believe that is not a simple task, but could possibly be achieved by reparameterization).
library(nlme)
library(reshape2)
df3 <- melt(df, id.vars="A")
r <- nlme(value ~ k1/A + k2,
data = df3,
start = c(k1=10, k2=10),
fixed = k1 + k2 ~1,
random = k2 ~ 1|variable)
summary(r)
# Nonlinear mixed-effects model fit by maximum likelihood
# Model: value ~ k1/A + k2
# Data: df3
# AIC BIC logLik
# 83.11052 84.32086 -37.55526
#
# Random effects:
# Formula: k2 ~ 1 | variable
# k2 Residual
# StdDev: 12.49915 7.991013
#
# Fixed effects: k1 + k2 ~ 1
# Value Std.Error DF t-value p-value
# k1 124.81916 9.737738 7 12.818086 0.0000
# k2 1.19925 11.198211 7 0.107093 0.9177
# Correlation:
# k1
# k2 -0.397
#
# Standardized Within-Group Residuals:
# Min Q1 Med Q3 Max
# -1.7520706 -0.5273469 0.2746039 0.5235343 1.4971808
#
# Number of Observations: 10
# Number of Groups: 2
coef(r)
# k1 k2
# B 124.8192 -10.81835
# C 124.8192 13.21684